1. (a) Define trigonometric equation with example.[br]Solution:[br]An equation involving trigonometric ratios is called trigonometric equation.[br]For example: [br](i) [math] \cos \theta = \frac{3}{2} [/math] [br](ii) [math] \sin^2 \beta = 2 [/math] [br][br](b) What do you mean by root (solution) of the given trigonometric equation?[br]Solution:[br]The value of an unknown angle which satisfies the given trigonometric equation is known as its root or solution.[br][br]2. (a) If [math] \sec \theta = -2 [/math] , what is the least positive angle in [math] 1^{st} [/math] quadrant.[br]Solution:[br]In first quadrant all the trigonometric ratios are positive, so there is not any angle in the first quadrant that satisfies [math] \sec \theta = -2 [/math] .[br][br]2. (b) How to find the angle in [math] 4^{th} [/math] quadrant, if the least positive angle [math] \theta [/math] is given.[br]Solution:[br]If the least positive angle [math] \theta [/math] is given, then the angle in [math] 4^{th} [/math] quadrant [math] = 360^{\circ} - \theta [/math] [br][br]2. (c) What are the minimum and maximum values of [math] \sin \theta [/math] and [math] \cos \theta [/math] [br]Solution:[br]The minimum value of [math] \sin \theta [/math] and [math] \cos \theta [/math] [math] = -1 [/math] [br]The maximum values of [math] \sin \theta [/math] and [math] \cos \theta [/math] [math] = 1 [/math] [br][br]3. Solve: [math] ( 0^{\circ} \le \theta \le 90^{\circ} ) [/math] [br](a) [math] \sin \theta = \frac{\sqrt{3}}{2} [/math] [br]Solution:[br][math] \begin{align} \sin \theta & = \frac{\sqrt{3}}{2} \\ [br]\text{or, } \sin \theta & = \sin 60^{\circ} \\[br]\therefore & \theta = 60^{\circ} \\[br]\end{align} [/math][br][br](b) [math] \cos \theta = \frac{ 1 }{2} [/math] [br]Solution:[br][math] \begin{align} \cos \theta & = \frac{1 }{2} \\ [br]\text{or, } \cos \theta & = \cos 60^{\circ} \\[br]\therefore \ \ & \theta = 60^{\circ} \\[br]\end{align} [/math][br][br](c) [math] \sqrt{3}\cot \theta = 1 [/math] [br]Solution:[br][math] \begin{align} \sqrt{3}\cot \theta & = 1 \\ [br]\text{or, } \cot \theta & = \frac{1}{\sqrt{3}} \\[br]\text{or, } \tan \theta & = \sqrt{3} \\[br]\text{or, } \tan \theta & = \tan 60^{\circ} \\[br]\therefore \ \ \theta & = 60^{\circ} \\[br]\end{align} [/math][br][br](d) [math] \tan \theta - 1 = 0 [/math] [br]Solution:[br][math] \begin{align} \tan \theta - 1 & = 0 \\ [br]\text{or, } \tan \theta & = 1 \\[br]\text{or, } \tan \theta & = \tan 45^{\circ} \\[br]\therefore \ \ \theta & = 45^{\circ} \\[br]\end{align} [/math][br][br](e) [math] 2\sin \theta - 1 = 0 [/math] [br]Solution:[br][math] \begin{align} 2\sin \theta - 1 & = 0 \\ [br]\text{or, } 2 \sin \theta & = 1 \\[br]\text{or, } \sin \theta & = \frac{1}{2} \\[br]\text{or, } \sin \theta & = \sin 30^{\circ} \\[br]\therefore \ \ \theta & = 30^{\circ} \\[br]\end{align} [/math][br][br](f ) [math] \sin \theta = 1[/math] [br]Solution:[br][math] \begin{align} \sin \theta & = 1 \\ [br]\text{or, } \sin \theta & = \sin 90^{\circ} \\[br]\therefore \ \ \theta & = 90^{\circ} \\[br]\end{align} [/math][br][br](g) [math] \cos \theta - \frac{1}{\sqrt{2}} = 0 [/math] [br]Solution:[br][math] \begin{align} \cos \theta - \frac{1}{\sqrt{2}} & = 0 \\ [br]\text{or, } \cos \theta & = \frac{1}{\sqrt{2}} \\[br]\text{or, } \cos \theta & = \cos 45^{\circ} \\[br]\therefore \ \ \theta & = 45^{\circ} \\[br]\end{align} [/math][br][br](h) [math] \sec \theta = 2 [/math] [br]Solution:[br][math] \begin{align} \sec \theta & = 2 \\ [br]\text{or, } \cos \theta & = \frac{1}{2} \\[br]\text{or, } \cos \theta & = \cos 30^{\circ} \\[br]\therefore \ \ \theta & = 30^{\circ} \\[br]\end{align} [/math][br][br]4. Solve: [math] ( 0^{\circ} \le \alpha, \theta \le 180^{\circ} ) [/math] [br](a) [math] 2\cos \theta + 1 = 0 [/math] [br]Solution:[br][math] \begin{align} 2\cos \theta + 1 & = 0 \\ [br]\text{or, } 2 \cos \theta & = -1 \\[br]\text{or, } \cos \theta & = \frac{-1}{2} \\[br]\text{or, } \cos \theta & = \cos ( 180^{\circ} -60^{\circ} ) \\[br]\text{or, } \cos \theta & = \cos 150^{\circ} \\[br]\therefore \ \ \theta & = 150^{\circ} \\[br]\end{align} [/math][br][br](b) [math] \sqrt{2} \sec \theta + 2 = 0 [/math] [br]Solution:[br][math] \begin{align} \sqrt{2} \sec \theta + 2 & = 0 \\ [br]\text{or, } \sqrt{2} \sec \theta & = -2 \\[br]\text{or, } \sec \theta & = \frac{-2}{\sqrt{2} } \\[br]\text{or, } \cos \theta & = - \frac{\sqrt{2}}{2} \\[br]\text{or, } \cos \theta & = -\frac{\sqrt{2}}{\sqrt{2} \times \sqrt{2} } \\[br]\text{or, } \cos \theta & = -\frac{1 }{ \sqrt{2} }\\[br]\text{or, } \cos \theta & = \cos (180^{\circ} - 45^{\circ}) \\[br]\therefore \ \ \theta & = 135^{\circ} \\[br]\end{align} [/math][br][br](c) [math] 2\sin \theta - \sqrt{3} = 0 [/math] [br]Solution:[br][math] \begin{align} 2\sin \theta - \sqrt{3} & = 0 \\ [br]\text{or, } 2\sin \theta & = \sqrt{3} \\[br]\text{or, } \sin \theta & = \frac{\sqrt{3}}{2} \\[br]\text{or, } \sin \theta & = \sin 60^{\circ} \\[br]\therefore \ \ \theta & = 60^{\circ} \\[br]\end{align} [/math][br][br](d) [math] 3\cot \theta - \sqrt{3} = 0 [/math] [br]Solution:[br][math] \begin{align} 3\cot \theta - \sqrt{3} & = 0 \\ [br]\text{or, } 3\cot \theta & = \sqrt{3} \\[br]\text{or, } \cot \theta & = \frac{\sqrt{3}}{3} \\[br]\text{or, } \cot \theta & = \frac{\sqrt{3}}{\sqrt{3} \times \sqrt{3} } \\[br]\text{or, } \cot \theta & = \frac{1}{ \sqrt{3} } \\[br]\text{or, } \tan \theta & = \sqrt{3} \\[br]\text{or, } \tan \theta & = \tan 60^{\circ} \\[br]\therefore \ \ \theta & = 60^{\circ} \\[br]\end{align} [/math][br][br](e) [math] \sqrt{3} \text{cosec} \theta - 2 = 0 [/math] [br]Solution:[br][math] \begin{align} \ & \sqrt{3} \text{ cosec} \theta - 2 = 0 \\ [br]\text{or, } & \sqrt{3} \text{cosec} \theta = 2 \\[br]\text{or, } & \text{cosec} \theta = \frac{2}{\sqrt{3}} \\[br]\text{or, } & \sin \theta = \frac{\sqrt{3}}{2} \\[br]\text{or, } & \sin \theta = \sin 60^{\circ} , \sin ( 180^{\circ} -60^{\circ} ) \\[br]\text{or, } & \sin \theta = \sin 60^{\circ}, \sin 120^{\circ} \\[br]\therefore \ \ & \theta = 60^{\circ}, 120^{\circ} \\[br]\end{align} [/math][br][br](f) [math] \sqrt{3} \tan \theta + 1 = 0 [/math] [br]Solution:[br][math] \begin{align} \ & \ \sqrt{3} \tan \theta + 1 = 0 \\ [br]\text{or, } & \sqrt{3} \tan \theta = - 1 \\[br]\text{or, } & \tan \theta = \frac{-1}{\sqrt{3}} \\[br]\text{or, } & \tan \theta = \tan (180^{\circ} - 30^{\circ} ) \\[br]\text{or, } & \tan \theta = \tan 150^{\circ} \\[br]\therefore \ \ & \theta = 150^{\circ} \\[br]\end{align} [/math][br][br]5. (a) Solve: [math] 0^{\circ} \le \alpha \le 180^{\circ} [/math] [br][math] 2\sin^2 \alpha -1 = 0 [/math] [br]Solution:[br][math] \begin{align} 2\sin^2 \alpha -1 = 0 \\ [br]\text{or, } 2\sin^2 \alpha & = 1 \\[br]\text{or, } \sin^2 \alpha & = \frac{1}{2} \\[br]\text{or, } \sin \alpha & = \pm \sqrt{\frac{1}{2}} \\ [br]\text{or, } \sin \alpha & = \pm \frac{1}{\sqrt{2}} \\ [br]\end{align} [/math][br]Now,[br][math] \begin{tabular}{|l|l| }[br]\hline[br]\text{Taking positive} & \text{Taking negative} \\[br] \sin \alpha = \frac{1}{\sqrt{2}} & \sin \alpha = - \frac{1}{\sqrt{2}} \\[br] \text{or, }\sin \alpha = \sin 45^{\circ}, \sin ( 180 ^{\circ} - 45^{\circ} ) & \text{ Rejected } \\[br] \text{or, } \alpha = 45^{\circ}, 135^{\circ} & \\[br]\hline[br] \end{tabular} [/math][br][math] \therefore \alpha = 45^{\circ}, 135^{\circ} [/math] [br][br]5. (b) Solve: [math] 0^{\circ} \le \alpha \le 180^{\circ} [/math] [br][math] 4\sin \alpha = 3 \text{cosec} \alpha [/math] [br]Solution:[br][math] \begin{align} 4\sin \alpha & = 3 \text{cosec} \alpha \\ [br]\text{or, } 4 \sin \alpha & = 3\times \frac{1}{\sin \alpha } \\[br]\text{or, } 4 \sin^2 \alpha & = 3 \\[br]\text{or, } \sin^2 \alpha & = \frac{3 }{4} \\[br]\text{or, } \sin \alpha & = \pm \sqrt{\frac{3}{4}} \\ [br]\text{or, } \sin \alpha & = \pm \frac{\sqrt{3}}{2} \\ [br]\end{align} [/math][br]Now,[br][math] \begin{tabular}{|l|l| }[br]\hline[br]\text{Taking positive } & \text{ Taking negative } \\[br] \text{or, } \sin \alpha = \frac{\sqrt{3}}{2} & \sin \alpha = - \frac{\sqrt{3}}{2} \\[br] \text{or, } \sin \alpha = \sin 60^{\circ}, \sin ( 180 ^{\circ} - 60^{\circ} ) & Rejected \\[br]\text{or, } \alpha = 60^{\circ}, 120^{\circ} & \\[br]\hline[br] \end{tabular} [/math][br][math] \therefore \alpha = 60^{\circ}, 120^{\circ} [/math] [br][br]5. (c) Solve: [math] 0^{\circ} \le \alpha \le 180^{\circ} [/math] [br][math] \tan^2 \alpha - 1 = 2 [/math] [br]Solution:[br][math] \begin{align} \tan^2 \alpha - 1 & = 2 \\ [br]\text{or, } \tan^2 \alpha & = 2 +1 \\[br]\text{or, } \tan^2 \alpha & = 3 \\[br]\text{or, } \tan \alpha & = \pm \sqrt{3} \\[br]\end{align} [/math][br]Now,[br][math] \begin{tabular}{|l|l| }[br]\hline[br]\text{ Taking positive } & \text{ Taking negative } \\[br] \tan \alpha = \sqrt{3} & \tan \alpha = - \sqrt{3} \\[br] or, \tan \alpha = \tan 60^{\circ} & or, \tan \alpha = \tan ( 180 ^{\circ} - 60^{\circ} ) \\[br] or, \alpha = 60^{\circ} & or, \alpha = 120^{\circ} \\[br]\hline[br] \end{tabular} [/math][br][math] \therefore \alpha = 60^{\circ}, 120^{\circ} [/math] [br]5. (d) Solve: [math] 0^{\circ} \le \theta \le 180^{\circ} [/math] [br][math] 3\cot^2 \theta - 1 = 0 [/math] [br]Solution:[br][math] \begin{align} 3 \cot^2 \theta - 1 = 0 \\ [br]\text{or, } 3 \cot^2 \theta = 1 \\[br]\text{or, } \cot^2 \theta = \frac{1}{3} \\[br]\text{or, } \tan^2 \theta = 3 \\ [br]\text{or, } \tan \theta = \pm \sqrt{3} \\ [br]\end{align} [/math][br]Now,[br][math] \begin{tabular}{|l|l| }[br]\hline[br]\text{Taking positive} & \text{Taking negative} \\[br] \tan \alpha = \sqrt{3} & \tan \alpha = - \sqrt{3} \\[br]or, \tan \alpha = \tan 60^{\circ} & or, \tan \alpha = \tan ( 180 ^{\circ} - 60^{\circ} ) \\[br] or, \alpha = 60^{\circ} & or, \alpha = 120^{\circ} \\[br]\hline[br] \end{tabular} [/math] [br][math] \therefore \alpha = 60^{\circ}, 120^{\circ} [/math] [br][br]5. (e) Solve: [math] (0^{\circ}\le \alpha, \theta \le 180^{\circ} ) [/math] [math] 4\sin^2 \alpha = \tan^2 60^{\circ} [/math] [br]Solution:[br][math] \begin{align} 4\sin^2 \alpha & = \tan^2 60^{\circ} \\ [br]\text{or, } 4\sin^2 \alpha & = (\sqrt{3} )^2 \\[br]\text{or, } 4\sin^2 \alpha & = 3 \\[br]\text{or, } \sin^2 \alpha & = \frac{3}{4} \\[br]\text{or, } \sin \alpha & = \pm \sqrt{\frac{3}{4}} \\ [br]\text{or, } \sin \alpha & = \pm \frac{\sqrt{3}}{2} \\ [br]\end{align} [/math][br]Now,[br][math] \begin{tabular}{|l|l| }[br]\hline[br]\text{ Taking positive } & \text{ Taking negative } \\[br] \sin \alpha = \frac{\sqrt{3}}{2} & \sin \alpha = - \frac{\sqrt{3}}{2} \\[br] \sin \alpha = \sin 60^{\circ}, \sin ( 180 ^{\circ} - 60^{\circ} ) & \sin \alpha = \sin (180^{\circ}+60^{\circ} ), \sin (360^{\circ}-60^{\circ}) \\[br] \alpha = 60^{\circ}, 120^{\circ} & \alpha = 120^{\circ}, 300^{\circ} \\[br]\hline[br] \end{tabular} [/math][br][math] \therefore \alpha = 60^{\circ},120^{\circ},240^{\circ},300^{\circ} [/math] [br][br]5. (f) Solve: [math] 0^{\circ} \le \theta \le 180^{\circ} [/math] [br][math] \sqrt{3} \tan \theta + 1 = 0 [/math] [br]Solution:[br][math] \begin{align} & \sqrt{3} \tan \theta + 1 = 0 \\ [br]\text{or, } & \sqrt{3} \tan \theta = - 1 \\[br]\text{or, } & \tan \theta = \frac{-1}{\sqrt{3}} \\[br]\text{or, } & \tan \theta = \tan ( 180 ^{\circ} - 60^{\circ} ) \\[br]\text{or, } & \theta = 180 ^{\circ} - 60^{\circ} \\[br]\therefore \ \ & \ \ \theta = 120 ^{\circ} \\[br]\end{align} [/math][br][br]6. (a) Solve: [math] (0^{\circ}\le \theta \le 180^{\circ} ) [/math] [br][math] 2\cos^2 \theta = - \sqrt{3} \cos \theta [/math] [br]Solution:[br][math] \begin{align} \ & 2\cos^2 \theta = - \sqrt{3} \cos \theta \\ [br]\text{or, } & 2\cos^2 \theta + \sqrt{3} \cos \theta = 0 \\[br]\text{or, } & \cos \theta ( 2\cos \theta + \sqrt{3} ) = 0 \\[br]\end{align} [/math][br]Now,[br][math] \begin{tabular}{|l|l| }[br]\hline[br]\text{ Either} & \text{ Or } \\[br] \cos \theta = 0 & 2\cos \theta + \sqrt{3} = 0 \\[br]or, \cos \theta = \cos 90^{\circ} & or, 2\cos \theta = - \sqrt{3} \\[br]or, \theta = 90^{\circ} & or, \cos \theta = \frac{-\sqrt{3}}{2} \\[br] \ & or, \cos \theta = \cos (180^{\circ} - 30^{\circ} ) \\[br] \ & or, \cos \theta = \cos 150^{\circ} \\[br] \ & or, \theta = 150^{\circ} \\[br]\hline[br] \end{tabular} [/math] [br][math] \therefore \theta = 90^{\circ}, 150^{\circ} [/math][br][br]6. (b) Solve: [math] (0^{\circ}\le \theta \le 180^{\circ} ) [/math] [br][math] 2\cos^2 \theta =3\sin \theta [/math] [br]Solution:[br][math] \begin{align} \ & 2\cos^2 \theta =3\sin \theta \\ [br]\text{or, } & 2(1- \sin^2 \theta) = 3 \sin \theta \\[br]\text{or, } & 2 - 2\sin^2 \theta = 3 \sin \theta \\[br]\text{or, } & 0 = 2\sin^2 \theta + 3 \sin \theta - 2 \\[br]\text{or, } & 2\sin^2 \theta + 3 \sin \theta - 2 = 0 \\[br]\text{or, } & 2\sin^2 \theta + 4 \sin \theta -\sin \theta - 2 = 0 \\[br]\text{or, } & 2\sin \theta ( \sin \theta + 2 ) - 1( \sin \theta + 2 ) = 0 \\[br]\text{or, } & ( \sin \theta + 2 )( 2\sin \theta - 1 )= 0 \\[br]\end{align} [/math][br]Now,[br][math][br]\begin{tabular}{|l|l| }[br]\hline[br]\text{Either } & \text{ Or } \\[br] \sin \theta + 2 = 0 & 2\sin \theta - 1 \\[br] or, \sin \theta = - 2 & or, 2\sin \theta = 1 \\[br] \text{Rejected } & or, \sin \theta = \frac{1}{2} \\[br] \ & or, \sin \theta = \sin 30^{\circ} , \sin (180^{\circ} - 30^{\circ} ) \\[br] \ & or, \cos \theta = \sin 30^{\circ}, \sin 150^{\circ} \\[br] \ & or, \theta = 30^{\circ}, 150^{\circ} \\[br]\hline[br] \end{tabular} [/math][br][math] \therefore \ \ \theta = 30^{\circ}, 150^{\circ} [/math] [br][br]6. (c) Solve: [math] (0^{\circ}\le \theta \le 180^{\circ} ) [/math] [br][math] \text{cosec} \theta - 2\sin \theta = 1 [/math] [br]Solution:[br][math] \begin{align} \ & \text{cosec} \theta - 2\sin \theta = 1 \\ [br]\text{or, } & \frac{1}{\sin \theta } - 2\sin \theta = 1 \\[br]\text{or, } & \frac{1-2\sin^2 \theta}{\sin \theta } = 1\\[br]\text{or, } & 1-2\sin^2 \theta = \sin \theta \\[br]\text{or, } & 0 = 2\sin^2 \theta + \sin \theta -1 \\[br]\text{or, } & 2\sin^2 \theta + \sin \theta -1 = 0 \\[br]\text{or, } & 2\sin^2 \theta + 2 \sin \theta -\sin \theta - 1 = 0 \\[br]\text{or, } & 2\sin \theta ( \sin \theta + 1 ) - 1( \sin \theta + 1 ) = 0 \\[br]\text{or, } & ( \sin \theta + 1 )( 2\sin \theta - 1 )= 0 \\[br]\end{align} [/math][br]Now,[br][math][br]\begin{tabular}{|l|l| }[br]\hline[br]\text{Either } & \text{ Or } \\[br] \sin \theta +1 = 0 & 2\sin \theta - 1 \\[br] or, \sin \theta = - 1 & or, 2\sin \theta = 1 \\[br]\text{Rejected } & or, \sin \theta = \frac{1}{2} \\[br]\ & or, \sin \theta = \sin 30^{\circ} , \sin (180^{\circ} - 30^{\circ} ) \\[br] \ & or, \theta = 30^{\circ}, 150^{\circ} \\[br]\hline[br] \end{tabular} [/math][br][math] \therefore \ \ \theta = 30^{\circ}, 150^{\circ} [/math] [br][br]6. (d) Solve: [math] (0^{\circ}\le \theta \le 180^{\circ} ) [/math] [br][math] \cos^2 \frac{\theta}{2} -\cos \frac{\theta}{2} + \frac{1}{4} = 0 [/math] [br][br]Solution:[br][math] \begin{align} \ & \cos^2 \frac{\theta}{2} -\cos \frac{\theta}{2} + \frac{1}{4} = 0 \\ [br]\text{or, } & \frac{4 \cos^2 \frac{\theta}{2} - 4 \cos \frac{\theta}{2} + 1 }{4} = 0 \\[br]\text{or, } & 4 \cos^2 \frac{\theta}{2} - 4 \cos \frac{\theta}{2} + 1 =0 \\[br]\text{or, } & \left(2\cos \frac{\theta}{2} - 1 \right)^2 = 0 \\[br]\text{or, } & 2\cos \frac{\theta}{2} - 1 =0 \\[br]\text{or, } & 2\cos \frac{\theta}{2} = 1 \\[br]\text{or, } & \cos \frac{\theta}{2} = \frac{1}{2} \\[br]\text{or, } & \cos \frac{\theta}{2} = \cos 30^{\circ} \\[br]\text{or, } & \cos \frac{\theta}{2} = \cos 30^{\circ} \\[br]\text{or, } & \frac{\theta}{2} = 30^{\circ} \\[br]\text{or, } & \theta = 2\times 30^{\circ} \\[br]\therefore \ \ \ & \theta = 60^{\circ} \\[br]\end{align} [/math][br][br]6. (e) Solve: [math] (0^{\circ}\le \theta \le 180^{\circ} ) [/math] [br][math] \cos \theta ( 2\sin \theta - 1 ) = 0 [/math] [br]Solution:[br][math] \cos \theta ( 2\sin \theta - 1 ) = 0 [/math] [br]Now,[br][math][br]\begin{tabular}{|l|l| }[br]\hline[br]\text{Either} & \text{ Or } \\[br] \cos \theta = 0 & 2\sin \theta - 1 \\[br] or, \cos \theta = \cos 90^{\circ} & or, 2\sin \theta = 1 \\[br] or, \theta = 90^{\circ} & or, \sin \theta = \frac{1}{2} \\[br]\ & or, \sin \theta = \sin 30^{\circ} , \sin (180^{\circ} - 30^{\circ} ) \\[br] \ & or, \theta = 30^{\circ}, 150^{\circ} \\[br]\hline[br] \end{tabular} [/math][br][math] \therefore \theta = 30^{\circ}, 90^{\circ}, 150^{\circ} [/math] [br][br]6. (f) Solve: [math] (0^{\circ}\le \theta \le 180^{\circ} ) [/math] [br][math] \sin 2\theta = \sin \theta [/math] [br]Solution:[br][math] \begin{align} \ & \sin 2\theta = \sin \theta \\ [br]\text{or, } & 2\sin \theta \cos \theta = \sin \theta \\[br]\text{or, } & 2\sin \theta \cos \theta - \sin \theta =0 \\ [br]\text{or, } & \sin \theta (2 \cos \theta - 1)=0 \\ [br]\end{align} [/math][br]Now,[br][math][br]\begin{tabular}{|l|l| }[br]\hline[br]\text{Either} & \text{ Or } \\[br] \sin \theta = 0 & 2\cos \theta - 1 \\[br] or, \sin \theta = \sin 90^{\circ} & or, 2\cos \theta = 1 \\[br]or, \theta = 90^{\circ} & or, \cos \theta = \frac{1}{2} \\[br]\ & or, \cos \theta = \cos 60^{\circ} , \cos (180^{\circ} - 60^{\circ} ) \\[br] \ & or, \theta = 60^{\circ}, 120^{\circ} \\[br]\hline[br] \end{tabular} [/math][br][math] \therefore \theta = 60^{\circ}, 90^{\circ}, 120^{\circ} [/math] [br][br]6. (g) Solve: [math] (0^{\circ}\le \theta \le 180^{\circ} ) [/math] [br][math] \sin 2\theta = \cos 6\theta [/math] [br]Solution:[br][math] \begin{align} \sin 3\theta & = \cos 6\theta \\ [br]\text{or, } \sin 3\theta & = \sin (90^{\circ} - 6\theta ) \sin (5\times 90^{\circ} - 6\theta ), \sin (9\times 90^{\circ} - 6\theta ) , \sin (13\times 90^{\circ} - 6\theta ), \sin (17\times 90^{\circ} - 6\theta ) \\[br]\text{or, } 3\theta & = 90^{\circ} - 6\theta, 5\times 90^{\circ} - 6\theta, 9\times 90^{\circ} - 6\theta,13\times 90^{\circ} - 6\theta,17\times 90^{\circ} - 6\theta \\[br]\text{or, } 3\theta + 6\theta & = 90^{\circ} , 5\times 90^{\circ}, 9\times 90^{\circ} ,13\times 90^{\circ},17\times 90^{\circ}\\[br]\text{or, } 9\theta & = 90^{\circ} , 5\times 90^{\circ}, 9\times 90^{\circ} ,13\times 90^{\circ},17\times 90^{\circ}\\[br]\text{or, } 9\theta & = 90^{\circ} , 5\times 90^{\circ}, 9\times 90^{\circ} ,13\times 90^{\circ},17\times 90^{\circ}\\[br]\text{or, } \theta & = \frac{90^{\circ}}{9} , \frac{5\times 90^{\circ}}{9}, \frac{9\times 90^{\circ}}{9} ,\frac{13\times 90^{\circ}}{9},\frac{17\times 90^{\circ}}{9}\\[br]\text{or, } \theta & = 10^{\circ}, 50^{\circ}, 90^{\circ}, 130^{\circ}, 170^{\circ} \\[br]\end{align} [/math][br][br]6. (h) Solve: [math] (0^{\circ}\le \theta \le 180^{\circ} ) [/math] [br][math] \cot 5\theta = \tan \theta [/math] [br]Solution:[br][math] \begin{align} \cot 5\theta & = \tan \theta \\ [br]\text{or, } \cot 5\theta & = \cot (90^{\circ} - \theta ) \sin (3\times 90^{\circ} - \theta ), \sin (5\times 90^{\circ} - \theta ) , \sin (7\times 90^{\circ} - \theta ), \sin (9\times 90^{\circ} - \theta ), \sin (11\times 90^{\circ} - \theta ) \\[br]\text{or, } 5 \theta & = 90^{\circ} - \theta, 3\times 90^{\circ} - \theta, 5\times 90^{\circ} - \theta,7\times 90^{\circ} - \theta,9\times 90^{\circ} - \theta, 11\times 90^{\circ} - \theta \\[br]\text{or, } 5\theta + \theta & = 90^{\circ} , 3\times 90^{\circ}, 5\times 90^{\circ} ,7\times 90^{\circ},9\times 90^{\circ},11\times 90^{\circ} - \theta\\[br]\text{or, } 6\theta & = 90^{\circ} , 3\times 90^{\circ}, 5\times 90^{\circ} ,7\times 90^{\circ},9\times 90^{\circ},11\times 90^{\circ} \\[br]\text{or, } 6\theta & = 90^{\circ} , 3\times 90^{\circ}, 5\times 90^{\circ} ,7\times 90^{\circ},9\times 90^{\circ},11\times 90^{\circ}\\[br]\text{or, } \theta & = \frac{90^{\circ}}{6} , \frac{3\times 90^{\circ}}{6}, \frac{5\times 90^{\circ}}{6} ,\frac{7\times 90^{\circ}}{6},\frac{9\times 90^{\circ}}{6}, \frac{11\times 90^{\circ}}{6}\\[br]\text{or, } \theta & = 15^{\circ}, 45^{\circ}, 75^{\circ},105^{\circ},135^{\circ},165^{\circ} \\[br]\end{align} [/math][br][br]7. Solve:- [math] ( 0^{\circ} \le x \le 360^{\circ} ) [/math] [br][br](a) [math] 3\sin^2 x + 4\cos x = 4 [/math] [br]Solution:[br][math] \begin{align} [br]\ & 3\sin^2 x + 4\cos x = 4 \\[br]\text{or, } & 3(1-\cos^2 x ) + 4\cos x = 4 \\[br]\text{or, } & 3 - 3\cos^2 x + 4\cos x = 4 \\[br]\text{or, } & 0 = 3\cos^2 x - 4\cos x + 4 - 3 \\[br]\text{or, } & 3\cos^2 x - 4\cos x + 1 = 0 \\[br]\text{or, } & 3\cos^2 x - (3+1)\cos x + 1 = 0 \\[br]\text{or, } & 3\cos^2 x - 3\cos x - \cos x+ 1 = 0 \\[br]\text{or, } & 3\cos x ( \cos x - 1 ) - 1 ( \cos x - 1 ) = 0 \\[br]\text{or, } & (\cos x - 1 )(3\cos x - 1 ) = 0 \\[br]\end{align} [/math][br][math] \begin{tabular}{|l|l| }[br]\hline[br]\text{Either} & \text{Or} \\[br] \cos x - 1 = 0 & 3\cos x - 1 = 0 \\[br]or, \cos x = 1 & or, 3\cos x = 1 \\[br]or, \cos x = \cos 0^{\circ}, \cos 360^{\circ} & or, \cos x = \frac{1}{3} \\[br]or, x = 0^{\circ} , 360^{\circ} \ & or, x = \cos^{-1}\left(\frac{1}{3} \right) \\[br]\hline[br] \end{tabular} [/math][br][math] \therefore x = 0^{\circ}, 360^{\circ}, \cos^{-1}\left( \frac{1}{3} \right) [/math] [br][br]7. Solve:- [math] ( 0^{\circ} \le x \le 360^{\circ} ) [/math] [br](b) [math] \cos^2 x = 3\sin^2 x + 4\cos x [/math] [br]Solution:[br][math] \begin{align} [br]\ & \cos^2 x = 3\sin^2 x + 4\cos x \\[br]\text{or, } & \cos^2 x = 3( 1 - \cos^2 x ) + 4\cos x \\[br]\text{or, } & \cos^2 x = 3 - 3\cos^2 x + 4\cos x \\[br]\text{or, } & \cos^2 x + 3\cos^2 x + 4\cos x - 3 = 0\\[br]\text{or, } & 4\cos^2 x + 4\cos x - 3 = 0 \\[br]\text{or, } & 4\cos^2 x + (6-2)\cos x - 3 = 0 \\[br]\text{or, } & 4\cos^2 x + 6\cos x-2\cos x - 3 = 0 \\[br]\text{or, } & 2\cos x (2\cos x + 3) -1 (2\cos x + 3) = 0 \\[br]\text{or, } & (2\cos x + 3) (2\cos x - 1 ) = 0 \\[br]\end{align} [/math][br][math] \begin{tabular}{|l|l| }[br]\hline[br]\text{Either} & \text{Or } \\[br] 2\cos x + 3 =0 & 2\cos x - 1= 0 \\[br]or, 2\cos x = -3 & or, 2\cos x = 1 \\[br]Rejected & or, \cos x = \frac{1}{2} \\[br] \ & or, \cos x = \cos 60^{\circ}, \cos( 360^{\circ} - 60^{\circ} ) \\[br] \ & or, \cos x = \cos 60^{\circ}, \cos 300^{\circ} \\[br]\ & or, x = 60^{\circ} , 300^{\circ} \\[br]\hline[br] \end{tabular} [/math][br][math] \therefore x = 60^{\circ}, 30^{\circ} [/math] [br][br]7. Solve:- [math] ( 0^{\circ} \le x \le 360^{\circ} ) [/math] [br](c) [math] \tan x + \cot x = 2 [/math] [br][math] \begin{align} [br]\ & \tan x + \cot x = 2 \\[br]\text{or, } & \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x } = 2 \\[br]\text{or, } & \frac{\sin^2 x + \cos^2 x }{\cos x \sin x } = 2 \\[br]\text{or, } & \frac{1}{\cos x \sin x } = 2 \\[br]\text{or, } & \frac{1}{\cos x \sin x } = 2 \\[br]\text{or, } & 1= 2 \sin x \cos x \\[br]\text{or, } & 2\sin x \cos x = 1 \\[br]\text{or, } & \sin 2x = 1 \\[br]\text{or, } & \sin 2x = \sin 90^{\circ}, \sin 5\times 90^{\circ} \\ [br]\text{or, } & 2x = 90^{\circ}, 5\times 90^{\circ} \\[br]\text{or, } & x = \frac{90^{\circ}}{2}, \frac{5\times 90^{\circ}}{2} \\[br]\therefore \ \ & x = 45^{\circ}, 225^{\circ} \\[br]\end{align} [/math][br][br]7. Solve:- [math] ( 0^{\circ} \le x \le 360^{\circ} ) [/math] [br](d) [math] \tan x -\sin x = 0 [/math] [br]Solution:[br][math] \begin{align} [br]\ & \tan x -\sin x = 0 \\[br]\text{or, } & \frac{\sin x }{\cos x} - \sin x = 0 \\[br]\text{or, } & \frac{\sin x - \sin x \cos x }{\cos x} = 0 \\[br]\text{or, } & \sin x - \sin x \cos x = 0 \\[br]\text{or, } & \sin x ( 1 - \cos x ) = 0 \\[br]\end{align} [/math][br][math] \begin{tabular}{|l|l| }[br]\hline[br]\text{Either} & \text{Or} \\[br]\sin x = 0 & 1 - \cos x = 0 \\[br]or, \sin x = \sin 90^{\circ} & or, \cos x = 1 \\[br] or, x = 90^{\circ} & or, \cos x = \cos 0^{\circ} \\[br] \ & or, \cos x = \cos 0^{\circ}, \cos( 360^{\circ} - 0^{\circ} ) \\[br] \ & or, \cos x = \cos 0^{\circ}, \cos 360^{\circ} \\[br]\ & or, x = 0^{\circ} , 360^{\circ} \\[br]\hline[br] \end{tabular} [/math][br][math] \therefore x = 0^{\circ}, 90^{\circ}, 360^{\circ} [/math] [br][br]7. Solve:- [math] ( 0^{\circ} \le x \le 360^{\circ} ) [/math] [br][br](e) [math] \sec x . \tan x = \sqrt{2} [/math] [br]Solution:[br][math] \begin{align} [br]\ & \sec x . \tan x = \sqrt{2} \\[br]\text{or, } & \frac{ 1 }{\cos x} . \frac{\sin x }{\cos x } =\sqrt{2} \\[br]\text{or, } & \frac{\sin x}{\cos^2 x } = \sqrt{2} \\[br]\text{or, } & \sin x = \sqrt{2} \cos^2 x \\[br]\text{or, } & \sin x = \sqrt{2} (1-\sin^2 x ) \\[br]\text{or, } & \sin x = \sqrt{2} - \sqrt{2} \sin^2 x \\[br]\text{or, } & \sqrt{2} \sin^2 x +\sin x - \sqrt{2} = 0 \\[br]\text{or, } & \sqrt{2} \sin^2 x +(2-1) \sin x - \sqrt{2} = 0 \\[br]\text{or, } & \sqrt{2} \sin^2 x +2 \sin x - \sin x - \sqrt{2} = 0 \\[br]\text{or, } & \sqrt{2} \sin x ( \sin x +\sqrt{2} ) - 1 (\sin x + \sqrt{2}) = 0 \\[br]\text{or, } & ( \sin x +\sqrt{2} ) (\sqrt{2} \sin x - 1 ) = 0 \\[br]\end{align} [/math][br][math] \begin{tabular}{|l|l| }[br]\hline[br]\text{Either} & \text{Or} \\[br] \sin x + \sqrt{2} = 0 & \sqrt{2} \sin x - 1 = 0 \\[br] or, \sin x = \sqrt{2} & or, \sqrt{2} \sin x =1 \\[br]\text{Rejected } & or, \sin x = \frac{1}{\sqrt{2}} \\[br]\ & or, \sin x = \sin 45^{\circ}, \sin (180^{\circ} - 45^{\circ} ) \\[br]\ & or, \sin x = \sin 45^{\circ}, \sin 135^{\circ} \\[br]\ & or, x = 45^{\circ} , 135^{\circ} \\[br]\hline[br] \end{tabular} [/math] [br][math] \therefore x = 45^{\circ} , 135^{\circ} [/math] [br][br]7. Solve:- [math] ( 0^{\circ} \le x \le 360^{\circ} ) [/math] [br](f) [math] \cot^2 x + \text{cosec}^2x = 3 [/math] [br]Solution:[br][math] \begin{align} [br]\ & \cot^2 x + \text{cosec}^2x = 3 \\[br]\text{or, } & \cot^2 x + 1+\cot^2 x = 3\\[br]\text{or, } & 2\cot^2 x = 3 - 1 \\[br]\text{or, } & 2\cot^2 x = 2 \\[br]\text{or, } & \cot^2 x = 1 \\[br]\text{or, } & \cot x = \pm \sqrt{ 1} \\[br]\text{or, } & \cot x = \pm 1 \\[br]\end{align} [/math][br][math] \begin{tabular}{|l|l| }[br]\hline[br]\text{Taking positive } & \text{ Taking negative } \\[br] \cot x = 1 & \cot x = -1 \\[br]or, \cot x = \cot 45^{\circ}, \cot (180^{\circ} + 45^{\circ} ) & or, \cot x = \cot( 180^{\circ} - 45^{\circ} ), \cot ( 360^{\circ} - 45^{\circ} ) \\[br] or, \cot x = \cot 45^{\circ}, \cot 225^{\circ} & or, \cot x = \cot 135^{\circ} , \cot 315^{\circ} \\[br]or, x =45^{\circ}, 225^{\circ} & x = 135^{\circ}, 315^{\circ} \\[br]\hline[br] \end{tabular} [/math][br][math] \therefore x = 45^{\circ} , 135^{\circ} , 225^{\circ}, 315^{\circ} [/math] [br][br]7. Solve:- [math] ( 0^{\circ} \le x \le 360^{\circ} ) [/math] [br](g) [math] (1-\sqrt{3} ) \tan x + 1 + \sqrt{3} = \sqrt{3} \sec^2 \theta [/math] [br]Solution:[br][math] \begin{align} [br]\ & (1-\sqrt{3} ) \tan x + 1 + \sqrt{3} = \sqrt{3} \sec^2 x \\[br]\text{or, } & \tan x - \sqrt{3} \tan x + 1 + \sqrt{3} = \sqrt{3} (1+\tan^2 x) \\[br]\text{or, } & \tan x - \sqrt{3} \tan x + 1 + \sqrt{3} = \sqrt{3} + \sqrt{3} \tan^2 x \\[br]\text{or, } & 0 = \sqrt{3} \tan^2 x+ \sqrt{3} \tan x - \tan x - 1 \\[br]\text{or, } & \sqrt{3} \tan^2 x+ \sqrt{3} \tan x - \tan x - 1 = 0 \\[br]\text{or, } & \sqrt{3} \tan x (\tan x + 1 ) - 1(\tan x + 1) = 0 \\[br]\text{or, } & (\tan x + 1 ) (\sqrt{3} \tan x + 1 ) = 0 \\[br]\end{align} [/math][br][math] \begin{tabular}{|l|l| }[br]\hline[br]\text{Either } & \text{ Or } \\[br] \tan x + 1 = 0 & \sqrt{3} \tan x + 1 = 0 \\[br] \tan x = -1 & \sqrt{3} \tan x = - 1 \\[br]or, \tan x = \tan ( 180^{\circ} - 45^{\circ}), \tan ( 360^{\circ}- 45^{\circ}) & or, \tan x = \frac{-1}{\sqrt{3}} \\[br]or, \tan x = \tan 135^{\circ}, \tan 315^{\circ} & or, \tan x = \tan (180^{\circ} - 30^{\circ} ) \tan ( 360^{\circ} - 30^{\circ} ) \\[br] or, x =135^{\circ}, 315^{\circ} & x = 150^{\circ}, 330^{\circ} \\[br]\hline[br] \end{tabular} [/math] [br][math] \therefore x = 135^{\circ} , 150^{\circ} , 315^{\circ}, 330^{\circ} [/math] [br][br]7. Solve:- [math] ( 0^{\circ} \le x \le 360^{\circ} ) [/math] [br](h) [math] \cot^2 x + \left( \sqrt{3} + \frac{1}{\sqrt{3}} \right) \cot x + 1 = 0 [/math] [br]Solution:[br][math] \begin{align} [br]\ & \cot^2 x + \left( \sqrt{3} + \frac{1}{\sqrt{3}} \right) \cot x + 1 = 0 \\[br]\text{or, } & \cot^2 x + \sqrt{3} \cot x + \frac{1}{\sqrt{3}} \cot x + 1 = 0 \\[br]\text{or, } & \cot x \left( \cot x + \sqrt{3} \right) + \frac{ 1}{\sqrt{3}} \left( \cot x + \sqrt{3} \right) = 0 \\[br]\text{or, } & \left( \cot x + \sqrt{3} \right) \left( \cot x + \frac{1}{\sqrt{3}} \right) = 0 \\[br]\end{align} [/math][br][math] \begin{tabular}{|l|l| }[br]\hline[br]\text{Either } & \text{ Or } \\[br] \cot x + \sqrt{3} = 0 & \cot x + \frac{1}{\sqrt{3}} = 0 \\[br] or, \cot x = -\sqrt{3} & or, \cot x = -\frac{1}{\sqrt{3}} \\[br]or, \cot x = \cot (180^{\circ}-30^{\circ}), \cot ( 360^{\circ} - 30^{\circ}) & \cot x = \cot ( 180^{\circ} - 60^{\circ} ) \cot ( 360^{\circ} - 60^{\circ} ) \\[br] or, \cot x = \cot 150^{\circ}, \cot 330^{\circ} & or, \cot x = \cot 120^{\circ}, \cot 300^{\circ} \\[br]or, x = 150^{\circ}, 330^{\circ} & or, x = 120^{\circ}, 300^{\circ} \\[br]\hline[br] \end{tabular} [/math] [br][math] \therefore x = 120^{\circ} , 150^{\circ} , 300^{\circ}, 330^{\circ} [/math] [br][br]7. Solve:- [math] ( 0^{\circ} \le x \le 360^{\circ} ) [/math] [br](i) [math] \tan^2 x + \left( 1- \sqrt{3} \right) \tan x - \sqrt{3} = 0 [/math] [br]Solution:[br][math] \begin{align} [br]\ & \tan^2 x + \left( 1- \sqrt{3} \right) \tan x - \sqrt{3} = 0 \\[br]\text{or, } & \tan^2 x + tan x - \sqrt{3} \tan x - \sqrt{3} = 0 \\[br]\text{or, } & \tan x ( \tan x + 1 ) - \sqrt{3} ( \tan x + 1 ) = 0 \\[br]\text{or, } & (\tan x + 1 ) ( \tan x - \sqrt{3} )= 0 \\[br]\end{align} [/math][br][math] \begin{tabular}{|l|l| }[br]\hline[br]\text{Either } & \text{ Or } \\[br] \tan x + 1 = 0 & \tan x - \sqrt{3} = 0 \\[br] or, \tan x = -1 & or, \tan x = \sqrt{3} \\[br] or, \tan x = \tan (180^{\circ}-45^{\circ}), \tan ( 360^{\circ} - 45^{\circ}) & or, \tan x = \tan ( 180^{\circ} - 60^{\circ} ) \tan ( 360^{\circ} - 60^{\circ} ) \\[br] or, \tan x = \tan 135^{\circ}, \tan 315^{\circ} & or, \tan x = \tan 120^{\circ}, \tan 300^{\circ} \\[br]or, x = 135^{\circ}, 315^{\circ} & or, x = 120^{\circ}, 300^{\circ} \\[br]\hline[br] \end{tabular} [/math][br][math] \therefore x = 120^{\circ} , 135^{\circ} , 300^{\circ}, 315^{\circ} [/math] [br][br]7. Solve:- [math] ( 0^{\circ} \le x \le 360^{\circ} ) [/math] [br](j) [math] 2\sin x + \cot x - \text{cosec} x = 0 [/math] [br]Solution:[br][math] \begin{align} [br]\ & 2\sin x + \cot x - \text{cosec} x = 0 \\[br]\text{or, } & 2\sin x + \frac{\cos x }{\sin x} -\frac{1}{ \sin x } = 0 \\[br]\text{or, } & \frac{ 2\sin^2 x + \cos x - 1 }{\sin x } = 0 \\[br]\text{or, } & 2\sin^2 x + \cos x - 1 = 0 \\[br]\text{or, } & 2(1-\cos^2 x )+ \cos x - 1 = 0 \\[br]\text{or, } & 2-2\cos^2 x + \cos x - 1 = 0 \\[br]\text{or, } & -2\cos^2 x + \cos x +1 = 0 \\[br]\text{or, } & 2\cos^2 x - \cos x - 1 = 0 \\[br]\text{or, } & 2\cos^2 x - (2-1) \cos x - 1 = 0 \\[br]\text{or, } & 2\cos^2 x - 2 \cos x + \cos x - 1 = 0 \\[br]\text{or, } & 2\cos x (2\cos x - 1 ) + 1 ( \cos x - 1 ) = 0 \\[br]\text{or, } & ( \cos x - 1 ) (2\cos x + 1 )= 0 \\[br]\end{align} [/math][br][math] \begin{tabular}{|l|l| }[br]\hline[br]\text{Either} & \text{ Or } \\[br] \cos x - 1 = 0 & 2\cos x + 1 = 0 \\[br] or, \cos x = 1 & or, 2\cos x = -1 \\[br] or, \cos x = \cos 0^{\circ}, \cos 360^{\circ} & or, \cos x = \frac{-1}{2} \\[br] or, x = 0^{\circ}, 360^{\circ} & or, \cos x = \cos (180^{\circ}-60^{\circ}), \cos ( 180^{\circ}+60^{\circ} ) \\[br]\ & \\ or, x = 120^{\circ}, 240^{\circ} \\[br]\hline[br] \end{tabular} [/math] [br][math] \therefore x = 0^{\circ}, 120^{\circ} , 240^{\circ} , 360^{\circ} [/math][br] [br]8. Solve:- [math] ( 0^{\circ} \le \theta \le 360^{\circ} ) [/math] [br](a) [math] \sqrt{3} \sin \theta + \cos \theta = 1 [/math] [br]Solution:[br][math] \sqrt{3} \sin \theta + \cos \theta = 1 ...(i) [/math] [br][math] \begin{align} [br]\ \text{Now} & \\[br] \ & \sqrt{(\text{coeff. of } \sin \theta )^2 + (\text{ coeff. of } \cos \theta )^2 } \\[br]\ & = \sqrt{ ( \sqrt{3} )^2 + 1^2 }\\ & = \sqrt{3+1}\\ & = \sqrt{4}\\ & = 2[br]\end{align} [/math][br][math] \begin{align} [br]\ & \text{ Dividing both sides of } eq^n(i) \text{ by } 2\\[br]\ & \text{We get, }\\[br]\ & \frac{\sqrt{3}}{2} \sin \theta + \frac{1}{2} \cos \theta = \frac{1}{2} \\[br]\text{or, } & \sin \theta \times \frac{\sqrt{3}}{2} + \cos \theta \times \frac{1}{2} = \frac{1}{2} \\[br]\text{or, } & \sin \theta \cos 30^{\circ} + \cos \theta \sin 30^{\circ} = \frac{1}{2} \\[br]\text{or, } & \sin ( \theta + 30^{\circ} ) = \sin 30^{\circ}, \sin ( 180^{\circ} - 30^{\circ} ), \sin ( 360^{\circ}‍‌‌‍+ ‌‍30^{\circ} ) \\[br]\text{or, } & \sin ( \theta + 30^{\circ} ) = \sin 30^{\circ}, \sin 150^{\circ}, \sin 390^{\circ} \\[br]\text{or, } & \theta + 30^{\circ} = 30^{\circ}, 150^{\circ}, 390^{\circ} \\[br]\text{or, } & \theta = 30^{\circ} - 30^{\circ}, 150^{\circ}- 30^{\circ}, 390^{\circ}- 30^{\circ} \\[br]\therefore \ \ & \theta = 0^{\circ}, 120^{\circ} ,360^{\circ}\\[br]\end{align} [/math][br][br]8. Solve:- [math] ( 0^{\circ} \le \theta \le 360^{\circ} ) [/math] [br](b) [math] \sin \theta + \sqrt{3} \cos \theta = 1 [/math] [br]Solution:[br][math] \sin \theta + \sqrt{3} \cos \theta = 1 ...(i) [/math] [br][math] \begin{align} [br]\ \text{Now} & \\[br] \ & \sqrt{(\text{coeff. of } \sin \theta )^2 + (\text{ coeff. of } \cos \theta )^2 } \\[br]\ & = \sqrt{ 1^2+( \sqrt{3} )^2 }\\ & = \sqrt{1+3}\\ & = \sqrt{4}\\ & = 2[br]\end{align} [/math][br][math] \begin{align} [br]\ & \text{ Dividing both sides of } eq^n(i) \text{ by } 2\\[br]\ & \text{We get, }\\[br]\ & \frac{1}{2} \sin \theta + \frac{\sqrt{3}}{2} \cos \theta = \frac{1}{2} \\[br]\text{or, } & \sin \theta \times \frac{1}{2} + \cos \theta \times \frac{\sqrt{3}}{2} = \frac{1}{2} \\[br]\text{or, } & \sin \theta \cos 60^{\circ} + \cos \theta \sin 60^{\circ} = \frac{1}{2} \\[br]\text{or, } & \sin ( \theta + 60^{\circ} ) = \sin 30^{\circ}, \sin ( 180^{\circ} - 30^{\circ} ), \sin ( 360^{\circ} + 30^{\circ} ) \\[br]\text{or, } & \sin ( \theta + 60^{\circ} ) = \sin 30^{\circ}, \sin 150^{\circ}, \sin 390^{\circ} \\[br]\text{or, } & \theta + 60^{\circ} = 30^{\circ}, 150^{\circ}, 390^{\circ} \\[br]\text{or, } & \theta = 30^{\circ} - 60^{\circ}, 150^{\circ}- 60^{\circ}, 390^{\circ}- 60^{\circ} \\[br]\text{or, } & \theta = -30^{\circ}\text{ (Rejected) }, 90^{\circ}, 330^{\circ} \\[br]\ \therefore & \theta = 90^{\circ}, 330^{\circ} \\[br]\end{align} [/math][br][br]8. Solve:- [math] ( 0^{\circ} \le \theta \le 360^{\circ} ) [/math] [br](c) [math] \sin \theta + \cos \theta = \sqrt{2} [/math] [br]Solution:[br][math] \sin \theta + \cos \theta = \sqrt{2} ...(i) [/math] [br][math] \begin{align} [br]\ \text{Now} & \\[br] \ & \sqrt{(\text{coeff. of } \sin \theta )^2 + (\text{ coeff. of } \cos \theta )^2 } \\[br]\ & = \sqrt{ 1^2+ 1^2 }\\ & = \sqrt{1+1}\\ & = \sqrt{2}\\[br]\end{align} [/math][br][math] \begin{align} [br]\ & \text{ Dividing both sides of } eq^n(i) \text{ by } \sqrt{2} \\[br]\ & \text{We get, }\\[br]\ & \frac{1}{\sqrt{2}} \sin \theta + \frac{1}{\sqrt{2}} \cos \theta = \frac{\sqrt{2}}{\sqrt{2}} \\[br]\text{or, } & \sin \theta \times \frac{1}{\sqrt{2}} + \cos \theta \times \frac{1}{\sqrt{2}} = 1 \\[br]\text{or, } & \sin \theta \cos 45^{\circ} + \cos \theta \sin 45^{\circ} = 1 \\[br]\text{or, } & \sin ( \theta + 45^{\circ} ) =\sin 90^{\circ} \\[br]\text{or, } & \theta + 45^{\circ} = 90^{\circ} \\[br]\text{or, } & \theta = 90^{\circ} - 45^{\circ} \\[br]\therefore \ \ & \theta = 45^{\circ} \\[br]\end{align} [/math][br][br]8. Solve:- [math] ( 0^{\circ} \le \theta \le 360^{\circ} ) [/math] [br](d) [math] \cos \theta + \frac{1}{\sqrt{3}} \sin \theta = 1 [/math] [br]Solution:[br][math] \begin{align} \ & \cos \theta + \frac{1}{\sqrt{3}} \sin \theta = 1 \\ \ & \text{or,} \frac{\sqrt{3} \cos \theta + \sin \theta}{\sqrt{3}}= 1 \\ \ & \text{or, } \sqrt{3} \cos \theta + \sin \theta = \sqrt{3} \\ & \text{or, } \sin \theta + \sqrt{3} \cos \theta = \sqrt{3} ... (i) [br]\end{align} [/math][br][math] \begin{align} [br]\ \text{Now} & \\[br] \ & \sqrt{(\text{coeff. of } \sin \theta )^2 + (\text{ coeff. of } \cos \theta )^2 } \\[br]\ & = \sqrt{ 1^2+( \sqrt{3} )^2 }\\ & = \sqrt{1+3}\\ & = \sqrt{4}\\ & = 2[br]\end{align} [/math][br][math] \begin{align} [br]\ & \text{ Dividing both sides of } eq^n(i) \text{ by } 2\\[br]\ & \text{We get, }\\[br]\ & \frac{1}{2} \sin \theta + \frac{\sqrt{3}}{2} \cos \theta = \frac{\sqrt{3}}{2} \\[br]\text{or, } & \sin \theta \times \frac{1}{2} + \cos \theta \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} \\[br]\text{or, } & \sin \theta \cos 60^{\circ} + \cos \theta \sin 60^{\circ} = \sin 60^{\circ}, \sin ( 180^{\circ} - 60^{\circ} ) ,\sin ( 360^{\circ} + 60^{\circ} ) \\[br]\text{or, } & \sin ( \theta + 60^{\circ} ) = \sin 60^{\circ}, \sin 120^{\circ}, \sin 420^{\circ}\\[br]\text{or, } & \theta + 60^{\circ} = 60^{\circ},120^{\circ}, 420^{\circ} \\[br]\text{or, } & \theta = 60^{\circ} - 60^{\circ}, 120^{\circ}-60,420^{\circ}-60^{\circ} \\[br]\therefore \ \ & \theta = 0^{\circ},60^{\circ},360^{\circ} \\[br]\end{align} [/math][br][br]8. Solve:- [math] ( 0^{\circ} \le \theta \le 360^{\circ} ) [/math] [br](e) [math] \sin \theta + \cos \theta = \frac{1}{\sqrt{2}} [/math] [br]Solution:[br][math] \begin{align} [br]\ \text{Now} & \\[br] \ & \sqrt{(\text{coeff. of } \sin \theta )^2 + (\text{ coeff. of } \cos \theta )^2 } \\[br]\ & = \sqrt{ 1^2+1^2 }\\ & = \sqrt{1+1}\\ & = \sqrt{2}[br]\end{align} [/math][br][math] \begin{align} [br]\ & \text{ Dividing both sides of } eq^n(i) \text{ by } \sqrt{2} \\[br]\ & \text{We get, }\\[br]\ & \frac{1}{\sqrt{2}} \sin \theta + \frac{1}{\sqrt{2}} \cos \theta = \frac{\frac{1}{\sqrt{2}}}{\sqrt{2}} \\[br]\text{or, } & \sin \theta \times \frac{1}{\sqrt{2}} + \cos \theta \times \frac{1}{\sqrt{2}} = \frac{1}{2} \\[br]\text{or, } & \sin \theta \cos 45^{\circ} + \cos \theta \sin 45^{\circ} = \sin 30^{\circ}, \sin ( 180^{\circ} - 30^{\circ} ) ,\sin ( 360^{\circ} + 30^{\circ} ) \\[br]\text{or, } & \sin ( \theta + 45^{\circ} ) = \sin 30^{\circ}, \sin 150^{\circ}, \sin 390^{\circ}\\[br]\text{or, } & \theta + 45^{\circ} = 30^{\circ},150^{\circ}, 390^{\circ} \\[br]\text{or, } & \theta = 30^{\circ} - 45^{\circ}, 150^{\circ}-45^{\circ},390^{\circ}-45^{\circ} \\[br]\text{or, } & \theta = -15^{\circ}\text{ (Refected) },105^{\circ},345^{\circ} \\[br]\therefore \ \ & \theta = 105^{\circ},345^{\circ} \\[br]\end{align} [/math][br][br]8. Solve:- [math] ( 0^{\circ} \le x \le 360^{\circ} ) [/math] [br](f) [math] \cos x + \sqrt{3} \sin x = 2 [/math] [br]Solution:[br][math] \cos x + \sqrt{3} \sin x = 2 [/math] [br][math] \text{or, } \sqrt{3} \sin x + \cos x = 2 [/math] [br][math] \begin{align} [br]\ \text{Now} & \\[br] \ & \sqrt{(\text{coeff. of } \sin x )^2 + (\text{ coeff. of } \cos x )^2 } \\[br]\ & = \sqrt{ ( \sqrt{3} )^2 + 1^2 }\\ & = \sqrt{3+1}\\ & = \sqrt{4}\\ & = 2[br]\end{align} [/math][br][math] \begin{align} [br]\ & \text{ Dividing both sides of } eq^n(i) \text{ by } 2\\[br]\ & \text{We get, }\\[br]\ & \frac{\sqrt{3}}{2} \sin x + \frac{1}{2} \cos x = \frac{2}{2} \\[br]\text{or, } & \sin x \times \frac{\sqrt{3}}{2} + \cos x \times \frac{1}{2} = 1 \\[br]\text{or, } & \sin x \cos 30^{\circ} + \cos x \sin 30^{\circ} = 1 \\[br]\text{or, } & \sin ( x + 30^{\circ} ) = \sin 90^{\circ}\\[br]\text{or, } & x+ 30^{\circ} = 90^{\circ} \\[br]\text{or, } & x = 90^{\circ} - 30^{\circ} \\[br]\therefore \ \ & x = 60^{\circ}\\[br]\end{align} [/math][br][br]8. Solve:- [math] ( 0^{\circ} \le \theta \le 360^{\circ} ) [/math] [br](g) [math] \tan \theta + \sqrt{3} \sec \theta = \sqrt{3} [/math] [br]Solution:[br][math] \begin{align} & \tan \theta + \sqrt{3} \sec \theta = \sqrt{3} \\[br]\text{or, } & \frac{\sin \theta }{\cos \theta} + \sqrt{3} \times \frac{1}{\cos \theta } = \sqrt{3} \\[br]\text{or, } & \frac{ \sin \theta + \sqrt{3} }{\cos \theta } = \sqrt{3} \\[br]\text{or, } & \sin \theta + \sqrt{3} = \sqrt{3} \cos \theta \\[br]\text{or, } & \sin \theta - \sqrt{3} \cos \theta = -\sqrt{3} .......(i)[br]\end{align} [/math][br][math] \begin{align} [br]\ \text{Now} & \\[br] \ & \sqrt{(\text{coeff. of } \sin \theta )^2 + (\text{ coeff. of } \cos \theta )^2 } \\[br]\ & = \sqrt{1^2+ ( \sqrt{3} )^2 }\\ & = \sqrt{1+3}\\ & = \sqrt{4}\\ & = 2[br]\end{align} [/math][br][math] \begin{align} [br]\ & \text{ Dividing both sides of } eq^n(i) \text{ by } 2\\[br]\ & \text{We get, }\\[br]\ & \frac{1}{2} \sin \theta - \frac{\sqrt{3}}{2} \cos \theta = \frac{-\sqrt{3}}{2} \\[br]\text{or, } & \sin \theta \times \frac{1}{2} - \cos \theta \times \frac{\sqrt{3}}{2} = \frac{-\sqrt{3}}{2} \\[br]\text{or, } & \sin \theta \cos 60^{\circ} - \cos \theta \sin 60^{\circ} = \frac{-\sqrt{3}}{2} \\[br]\text{or, } & \sin ( \theta - 60^{\circ} ) = \sin (180^{\circ}+60^{\circ}), \sin (360^{\circ}-60^{\circ} )\\[br]\text{or, } & \theta- 60^{\circ} = 240^{\circ}, 300^{\circ} \\ [br]\text{or, } & \theta = 240^{\circ} + 60^{\circ}, 300^{\circ}+60^{\circ} \\[br]\therefore \ \ & \theta = 300^{\circ}, 360^{\circ}\\[br]\end{align} [/math][br][br]8. Solve:- [math] ( 0^{\circ} \le \theta \le 360^{\circ} ) [/math] [br](h) [math] \text{cosec}\theta + \cot \theta = \sqrt{3} [/math] [br]Solution:[br][math] \begin{align} & \text{cosec}\theta + \cot \theta = \sqrt{3} \\[br]\text{or, } & \frac{1 }{\sin \theta} + \frac{\cos \theta }{\sin \theta } = \sqrt{3} \\[br]\text{or, } & \frac{ 1 + \cos \theta }{\sin \theta } = \sqrt{3} \\[br]\text{or, } & 1 + \cos \theta = \sqrt{3} \sin \theta \\[br]\text{or, } & 1 = \sqrt{3} \sin \theta - \cos \theta \\[br]\text{or, } & \sqrt{3} \sin \theta - \cos \theta = 1 ........(i) [br]\end{align} [/math][br][math] \begin{align} [br]\ \text{Now} & \\[br] \ & \sqrt{(\text{coeff. of } \sin \theta )^2 + (\text{ coeff. of } \cos \theta )^2 } \\[br]\ & = \sqrt{ (- \sqrt{3} )^2 +1^2 }\\ & = \sqrt{3+1}\\ & = \sqrt{4}\\ & = 2 [br]\end{align} [/math][br][math] \begin{align} [br]\ & \text{ Dividing both sides of } eq^n(i) \text{ by } 2\\[br]\ & \text{We get, }\\[br]\ & \frac{\sqrt{3}}{2} \sin \theta - \frac{1}{2} \cos \theta = \frac{1}{2} \\[br]\text{or, } & \sin \theta \times \frac{\sqrt{3}}{2} - \cos \theta \times \frac{1}{2} = \frac{1}{2} \\[br]\text{or, } & \sin \theta \cos 30^{\circ} - \cos \theta \sin 30^{\circ} = \frac{1}{2} \\[br]\text{or, } & \sin ( \theta - 30^{\circ} ) = \sin 30^{\circ}, \sin (180^{\circ}-30^{\circ}), \sin (360^{\circ} + 30^{\circ} )\\[br]\text{or, } & \theta- 30^{\circ} = 30^{\circ}, 150^{\circ}, 390^{\circ} \\ [br]\text{or, } & \theta = 30^{\circ}+ 30^{\circ}, 150^{\circ} + 30^{\circ}, 390^{\circ} + 30^{\circ} \\ [br]\text{or, } & \theta = 60^{\circ}, 180^{\circ}, 420^{\circ} \text{ (Rejected) }\\ [br]\therefore \ \ & \theta = 60^{\circ}, 180^{\circ} \\[br]\end{align} [/math][br][br]8. Solve:- [math] ( 0^{\circ} \le \theta \le 360^{\circ} ) [/math] [br](i) [math] \sqrt{3} \tan \theta + 1 = \sec \theta [/math] [br]Solution:[br][math] \begin{align} & \sqrt{3} \tan \theta + 1 = \sec \theta \\[br]\text{or, } & \sqrt{3} \times \frac{\sin \theta }{\cos \theta} + 1 = \frac{1}{\cos \theta } \\[br]\text{or, } & \frac{ \sqrt{3} \sin \theta + \cos \theta }{\cos \theta } = \frac{1}{\cos \theta } \\[br]\text{or, } & \sqrt{3} \sin \theta + \cos \theta = 1 ........(i) [br]\end{align} [/math][br][math] \begin{align} [br]\ \text{Now} & \\[br] \ & \sqrt{(\text{coeff. of } \sin \theta )^2 + (\text{ coeff. of } \cos \theta )^2 } \\[br]\ & = \sqrt{ (\sqrt{3} )^2 +1^2 }\\ & = \sqrt{3+1}\\ & = \sqrt{4}\\ & = 2 [br]\end{align} [/math][br][math] \begin{align} [br]\ & \text{ Dividing both sides of } eq^n(i) \text{ by } 2\\[br]\ & \text{We get, }\\[br]\ & \frac{\sqrt{3}}{2} \sin \theta - \frac{1}{2} \cos \theta = \frac{1}{2} \\[br]\text{or, } & \sin \theta \times \frac{\sqrt{3}}{2} + \cos \theta \times \frac{1}{2} = \frac{1}{2} \\[br]\text{or, } & \sin \theta \cos 30^{\circ} + \cos \theta \sin 30^{\circ} = \frac{1}{2} \\[br]\text{or, } & \sin ( \theta + 30^{\circ} ) = \sin 30^{\circ}, \sin (180^{\circ}-30^{\circ}), \sin (360^{\circ} + 30^{\circ} )\\[br]\text{or, } & \theta + 30^{\circ} = 30^{\circ}, 150^{\circ}, 390^{\circ} \\ [br]\text{or, } & \theta = 30^{\circ} - 30^{\circ}, 150^{\circ} - 30^{\circ}, 390^{\circ} - 30^{\circ} \\ [br]\therefore \ \ & \theta =0^{\circ}, 120^{\circ}, 360^{\circ} \\ [br]\end{align} [/math][br][br]9. Solve:- [math] ( 0^{\circ} \le \theta \le 180^{\circ} ) [/math] [br](a) [math] \sin 4\theta + \sin 2\theta = 0 [/math] [br]Solution:[br][math] \begin{align} & \sin 4\theta + \sin 2\theta = 0 \\[br]\text{or, } & 2\sin \left(\frac{4\theta + 2\theta }{2} \right)\cos \left(\frac{4\theta - 2\theta }{2} \right) = 0 \\[br]\text{or, } & 2\sin \left(\frac{6\theta }{2} \right)\cos \left(\frac{2\theta }{2} \right) = 0 \\[br]\text{or, } & 2\sin 3\theta \cos \theta = 0 \\[br]\text{or, } & \sin 3\theta \cos \theta = 0[br]\end{align} [/math][br][math] \begin{tabular} { | l | l | } [br]\hline [br]\text{Either} & \text{ Or } \\[br] \sin 3 \theta = 0 & \cos \theta = 0 \\[br]\text{or, } \sin 3 \theta = \sin 0^{\circ}, \sin 180^{\circ}, \sin 360^{\circ},sin 540^{\circ} & \text{or, } \cos \theta = \cos 90^{\circ} \\[br]\text{or, } 3\theta = 0^{\circ}, 180^{\circ}, 360^{\circ}, 540^{\circ} & \text{or, } \theta = 90^{\circ} \\[br]\text{or, } \theta = \frac{0^{\circ}}{3},\frac{180^{\circ}}{3},\frac{360^{\circ}}{3},\frac{540^{\circ}}{3} & \\[br]\text{or, } \theta = 0^{\circ}, 60^{\circ}, 120^{\circ}, 180^{\circ} & \\[br]\hline[br]\end{tabular}[br][/math] [br][math] \therefore \theta = 0^{\circ}, 60^{\circ}, 90^{\circ}, 120^{\circ}, 180^{\circ} [/math] [br][br]9. Solve:- [math] ( 0^{\circ} \le \theta \le 180^{\circ} ) [/math] [br](b) [math] \sin 3\theta + \sin 2\theta = \sin \theta [/math] [br]Solution:[br][math] \begin{align} \ & \sin 3\theta + \sin 2\theta = \sin \theta \\[br]\text{or, } & \sin 3\theta - \sin \theta = \sin 2\theta \\[br]\text{or, } & 2\cos \left(\frac{3\theta + \theta }{2} \right)\sin \left(\frac{3\theta - \theta }{2} \right) =\sin 2\theta \\[br]\text{or, } & 2\sin \left(\frac{4\theta }{2} \right)\sin \left(\frac{2\theta }{2} \right) =\sin 2\theta \\[br]\text{or, } & 2\sin 2\theta \cos \theta =\sin 2\theta \\[br]\text{or, } & 2\sin 2\theta \cos \theta - \sin 2\theta = 0 \\[br]\text{or, } & \sin 2\theta (2\cos \theta - 1) = 0 \\[br]\end{align} [/math][br][math] [br]\begin{tabular} { | l | l | } [br]\hline[br]\text{Either} & \text{Or }\\[br] \sin 2\theta = 0 & 2\cos \theta - 1 = 0 \\[br]\text{or, } \sin 2\theta = \sin 0^{\circ}, \sin 180^{\circ}, \sin 360^{\circ} & \text{or, } 2\cos \theta = 1 \\[br] \text{or, } 2 \theta = 0^{\circ}, 180^{\circ}, 360^{\circ} & \text{or, } \cos \theta = \frac{1}{2} \\[br] \text{or, } \theta = \frac{0^{\circ}}{2}, \frac{180^{\circ}}{2}, \frac{360^{\circ}}{2} & \text{or, } \cos \theta = \cos 60^{\circ} \\[br] \text{or, } \theta = 0^{\circ}, 90^{\circ}, 180^{\circ} & \text{or, } \theta = 60^{\circ} \\[br]\hline[br]\end{tabular}[br][/math] [br] [math] \therefore \ \ \theta = 0^{\circ}, 60^{\circ}, 90^{\circ}, 180^{\circ} [/math] [br][br]9. Solve:- [math] ( 0^{\circ} \le \theta \le 180^{\circ} ) [/math] [br](c) [math] \cos 3\theta + \cos \theta = \cos 2\theta [/math] [br]Solution:[br][math] \begin{align} & \cos 3\theta + \cos \theta = \cos 2\theta \\[br]\text{or, } & 2\cos \left(\frac{3\theta + \theta }{2} \right)\cos \left(\frac{3\theta - \theta }{2} \right) =\cos 2\theta \\[br]\text{or, } & 2\cos \left(\frac{4\theta }{2} \right)\cos \left(\frac{2\theta }{2} \right) =\cos 2\theta \\[br]\text{or, } & 2\cos 2\theta \cos \theta =\cos 2\theta \\[br]\text{or, } & 2\cos 2\theta \cos \theta - \cos 2\theta = 0 \\[br]\text{or, } & \cos 2\theta (2\cos \theta - 1) = 0 \\[br]\end{align} [/math][br][math] [br]\begin{tabular} { | l | l | } [br]\hline[br]\text{Either} & \text{Or } \\[br] \cos 2\theta = 0 & 2\cos \theta - 1 = 0 \\[br] \text{or, } \cos 2\theta = \cos 90^{\circ}, \cos (360^{\circ} - 90^{\circ}) & \text{or, } 2\cos \theta = 1 \\[br] \text{or, } 2 \theta = 90^{\circ}, 270^{\circ} & \text{or, } \cos \theta = \frac{1}{2} \\[br] \text{or, } \theta = \frac{90^{\circ}}{2}, \frac{270^{\circ}}{2} & \text{or, } \cos \theta = \cos 60^{\circ} \\[br] \text{or, } \theta = 45^{\circ}, 135^{\circ} & \text{or, } \theta = 60^{\circ} \\[br]\hline[br]\end{tabular}[br][/math] [br][math] \therefore \ \ \theta = 45^{\circ}, 60^{\circ}, 135^{\circ} [/math] [br][br]9. Solve:- [math] ( 0^{\circ} \le \theta \le 180^{\circ} ) [/math] [br](d) [math] \cos \theta + \cos 3\theta =- \cos 5\theta [/math] [br]Solution:[br][math] \begin{align} & \cos \theta + \cos 3\theta =- \cos 5\theta \\[br]\text{or, } & \cos 5\theta + \cos 3\theta + \cos \theta = 0 \\[br]\text{or, } & 2\cos \left(\frac{5\theta + 3\theta }{2} \right)\cos \left(\frac{5\theta - 3\theta }{2} \right) +\cos \theta = 0 \\[br]\text{or, } & 2\cos \left(\frac{8\theta }{2} \right)\cos \left(\frac{2\theta }{2} \right) +\cos \theta = 0 \\[br]\text{or, } & 2\cos 4\theta \cos \theta +\cos \theta = 0 \\[br]\text{or, } & \cos \theta (2\cos 4\theta + 1) = 0 \\[br]\end{align} [/math][br][math] [br]\begin{tabular} { | l | l | } [br]\hline[br]\text{Either} & \text{Or} \\[br] cos \theta =0 & 2\cos 4\theta + 1 = 0 \\[br] \text{or, } \cos \theta = \cos 90^{\circ}, \cos (360^{\circ} -90^{\circ} ) & \text{or, } 2\cos 4\theta = -1 \\[br] \text{or, } \theta = 90^{\circ} & \text{or, } \cos 4 \theta = \cos (180^{\circ} - 60^{\circ}), \cos (180^{\circ}+60^{\circ} ) \\[br]\ & \text{or, } 4\theta = 120^{\circ}, 240^{\circ} \\[br]\ & \text{or, } \theta = \frac{ 120^{\circ}}{4},\frac{ 240^{\circ}}{4} \\[br]\ & \text{or, } \theta = 30^{\circ}, 60^{\circ} \\[br]\hline [br]\end{tabular}[br][/math] [br][math] \therefore \ \ \theta = 30^{\circ}, 60^{\circ}, 90^{\circ} [/math] [br][br]9. Solve:- [math] ( 0^{\circ} \le \theta \le 180^{\circ} ) [/math] [br](e) [math] \cos 3\theta + \cos \theta = 2\cos \theta [/math] [br]Solution:[br][math] \begin{align} & \cos 3\theta + \cos \theta = 2\cos \theta \\[br]\text{or, } & 2\cos \left(\frac{3\theta + \theta }{2} \right)\cos \left(\frac{3\theta - \theta }{2} \right) = 2\cos \theta \\[br]\text{or, } & 2\cos \left(\frac{4\theta }{2} \right)\cos \left(\frac{2\theta }{2} \right) = 2\cos \theta \\[br]\text{or, } & 2\cos 2\theta \cos \theta - 2\cos \theta = 0 \\[br]\text{or, } & 2\cos \theta (\cos 2\theta - 1) = 0 \\[br]\text{or, } & \cos \theta (\cos 2\theta - 1) = 0 [br]\end{align} [/math][br][math] [br]\begin{tabular} { | l | l | } [br]\hline[br]\text{Either } & \text{ Or } \\[br] \cos \theta = \cos 90^{\circ} & \cos 2\theta - 1 = 0 \\[br] \text{or, } \theta = 90^{\circ} & \text{or, } \cos 2\theta = 1 \\[br]\ & \text{or, } \cos 2\theta = \cos 0^{\circ}, \cos 360^{\circ} \\[br]\ & \text{or, } 2\theta = 0^{\circ}, 360^{\circ} \\[br]\ & \text{or, } \theta = \frac{0^{\circ}}{2}, \frac{360^{\circ}}{2} \\[br]\ & \text{or, } \theta = 0^{\circ}, 180^{\circ} \\[br]\hline[br]\end{tabular}[br][/math] [br][math] \therefore \ \ \theta = 0^{\circ}, 90^{\circ}, 180^{\circ} [/math][br] [br]9. Solve:- [math] ( 0^{\circ} \le \theta \le 180^{\circ} ) [/math] [br](f) [math] \sin 2\theta + \sin 4\theta = \cos \theta + \cos 3\theta [/math] [br]Solution:[br][math][br]\begin{align} \ & \sin 2\theta + \sin 4\theta = \cos \theta + \cos 3\theta \\[br]\text{or, } & \sin 4\theta + \sin 2\theta = \cos 3\theta + \cos \theta \\[br]\text{or, } & 2\sin \left(\frac{4\theta + 2\theta }{2} \right)\cos \left(\frac{4\theta - 2\theta }{2} \right) = 2\cos \left(\frac{3\theta + \theta }{2} \right)\cos \left(\frac{3\theta - \theta }{2} \right) \\[br]\text{or, } & 2\sin \left(\frac{6\theta }{2} \right)\cos \left(\frac{2\theta }{2} \right) = 2\cos \left(\frac{4\theta }{2} \right)\cos \left(\frac{2\theta }{2} \right) \\[br]\text{or, } & 2\sin 3\theta \cos \theta = 2\cos 2\theta \cos \theta \\[br]\text{or, } & \sin 3\theta \cos \theta = \cos 2\theta \cos \theta \\[br]\text{or, } & \sin 3\theta \cos \theta - \cos 2\theta \cos \theta = 0 \\[br]\text{or, } & \cos \theta (\sin 3\theta - \cos 2\theta ) = 0 \\[br]\end{align} [/math][br][math] [br]\begin{tabular} { | l | l | } [br]\hline[br]\text{Either} & \text{ Or } \\[br] \cos \theta = 0 & \sin 3\theta - \cos 2\theta ) = 0 \\[br] \text{or, } \cos \theta = \cos 90^{\circ} & \text{or, } \sin 3\theta = \cos 2\theta \\[br] \text{or, } \theta = 90^{\circ} & \text{or, } \sin 3\theta = \sin ( 90^{\circ} - 2\theta ), \sin ( 5\times 90^{\circ} - 2\theta ) \\[br]\ & \text{or, } 3\theta + 2\theta = 90^{\circ}, 5\times 90^{\circ} \\[br]\ & \text{or, } 5\theta = 90^{\circ}, 5\times 90^{\circ} \\[br]\ & \text{or, } \theta =\frac{ 90^{\circ}}{5}, \frac{ 5\times 90^{\circ}}{5} \\[br]\ & \text{or, } \theta =18^{\circ}, 90^{\circ} \\[br]\hline[br]\end{tabular}[br][/math] [br][math] \therefore \ \ \theta =18^{\circ}, 90^{\circ} [/math] [br][br]9. Solve:- [math] ( 0^{\circ} \le \theta \le 180^{\circ} ) [/math] [br](g) [math] \cos \theta + \sin \theta = \cos 2\theta + \sin 2\theta [/math] [br]Solution:[br][math] \begin{align} & \cos \theta + \sin \theta = \cos 2\theta + \sin 2\theta \\[br]\text{or, } & \cos \theta - \cos 2\theta = \sin 2\theta - \sin \theta \\[br]\text{or, } & 2\sin \left(\frac{\theta + 2\theta }{2} \right)\sin \left(\frac{2\theta - \theta }{2} \right) = 2\cos \left(\frac{2\theta + \theta }{2} \right)\sin \left(\frac{2\theta - \theta }{2} \right) \\[br]\text{or, } & 2\sin \left(\frac{3\theta }{2} \right)\sin \left(\frac{\theta }{2} \right) = 2\cos \left(\frac{3\theta }{2} \right)\sin \left(\frac{\theta }{2} \right) \\[br]\text{or, } & \sin \left(\frac{3\theta }{2} \right)\sin \left(\frac{\theta }{2} \right) = \cos \left(\frac{3\theta }{2} \right)\sin \left(\frac{\theta }{2} \right) \\[br]\text{or, } & \sin \left(\frac{3\theta }{2} \right)\sin \left(\frac{\theta }{2} \right) - \cos \left(\frac{3\theta }{2} \right)\sin \left(\frac{\theta }{2} \right) = 0 \\[br]\text{or, } & \sin \frac{\theta}{2} \left[ \sin\frac{3\theta}{2} -\cos \frac{3\theta}{2} \right] = 0 \\[br]\end{align} [/math][br][math] [br]\begin{tabular} { | l | l | } [br]\hline[br]\text{Either} & \text{Or} \\[br] \sin \frac{\theta}{2} = 0 & \sin \frac{3\theta}{2} - \cos \frac{3\theta}{2} = 0 \\[br]\text{or, } \sin \frac{\theta}{2} = \sin 0^{\circ} & \text{or, } \sin \frac{3\theta}{2} = \cos \frac{3\theta}{2} \\[br] \text{or, } \frac{\theta}{2} =0 & \text{or, } \sin \frac{3\theta}{2} = \sin \left( 90^{\circ} - \frac{3\theta}{2} \right), \sin \left( 5\times 90^{\circ} - \frac{3\theta}{2} \right) \\[br] \text{or, } \theta = 0 & \text{or, } \frac{3\theta}{2} = 90^{\circ} - \frac{3\theta}{2} , 5\times 90^{\circ} - \frac{3\theta}{2} \\[br]\ & \text{or, } \frac{3\theta}{2} +\frac{3\theta}{2} = 90^{\circ}, 5 \times 90^{\circ} \\[br]\ & \text{or, } \frac{6\theta}{2} = 90^{\circ}, 5 \times 90^{\circ} \\[br]\ & \text{or, } 3\theta = 90^{\circ} , 5 \times 90^{\circ} \\[br]\ & \text{or, }\theta = \frac{90^{\circ}}{3}, \frac{5 \times 90^{\circ}}{3} \\[br]\ & \text{or, }\theta = 30^{\circ} , 150^{\circ} \\[br]\hline[br]\end{tabular}[br][/math] [br][br]10. Solve: [math] 0^{\circ} \le \alpha \le 90^{\circ} [/math] [br][math] \frac{\sqrt{3}}{\sin 2\alpha} + \frac{1}{\cos 2\alpha} = 4 [/math] [br]Solution:[br][math] \begin{align} [br]\ & \frac{\sqrt{3}}{\sin 2\alpha} + \frac{1}{\cos 2\alpha} = 4\\[br]\text{or, } & \frac{ \sqrt{3} \cos 2\alpha + \sin 2\alpha }{\sin 2\alpha \cos 2\alpha } = 4 \\[br]\text{or, } & \sqrt{3} \cos 2\alpha + \sin 2\alpha = 4 \sin 2\alpha \cos 2\alpha \\[br]\text{or, } & \sqrt{3} \cos 2\alpha + \sin 2\alpha = 2 ( 2 \sin 2\alpha \cos 2\alpha ) \\[br]\text{or, } & \sqrt{3} \cos 2\alpha + \sin 2\alpha = 2\sin 4\alpha \\[br]\text{or, } & \sin 2\alpha +\sqrt{3} \cos 2\alpha = 2\sin 4\alpha ......(i)[br]\end{align} [/math] [br][math] \begin{align} [br]\ \text{Now} & \\[br] \ & \sqrt{(\text{coeff. of } \sin 2\alpha )^2 + (\text{ coeff. of } \cos 2\alpha )^2 } \\[br]\ & = \sqrt{1^2+ ( \sqrt{3} )^2 }\\ & = \sqrt{1+3}\\ & = \sqrt{4}\\ & = 2[br]\end{align} [/math][br][math] \begin{align} [br]\ & \text{ Dividing both sides of } eq^n(i) \text{ by } 2\\[br]\ & \text{We get, }\\[br]\ & \frac{1}{2} \sin 2\alpha + \frac{\sqrt{3}}{2} \cos 2\alpha = \frac{2\sin 4\alpha }{2} \\[br]\text{or, } & \sin 2\alpha \times \frac{1}{2} + \cos 2\alpha \times \frac{\sqrt{3}}{2} = \sin 4\alpha \\[br]\text{or, } & \sin 2\alpha \cos 60^{\circ} + \cos 2\alpha \sin 60^{\circ} = \sin 4\alpha \\[br]\text{or, } & \sin ( 2\alpha + 60^{\circ} ) = \sin 4\alpha, \sin (180^{\circ} - 4\alpha), \sin (360^{\circ} + 4\alpha ) \\[br]\text{or, } & 2\alpha + 60^{\circ} = 4\alpha , 180^{\circ} - 4\alpha, 3\times 180^{\circ} - 4\alpha \\ [br]\end{align} [/math][br][math][br]\begin{tabular}{ |l |l |l | }[br]\hline[br]\text{Either} & \text{Or} & \text{Or } \\[br] 2\alpha + 60^{\circ} = 4 \alpha & 2\alpha + 60^{\circ} = 180^{\circ} - 4\alpha & 2\alpha + 60^{\circ} = 3\times 180^{\circ} - 4\alpha \\[br] \text{or, } 60^{\circ} = 4\alpha - 2\alpha & \text{or,} 2\alpha + 4\alpha = 180^{\circ} - 60^{\circ} & \text{or, } 2\alpha + 4\alpha = 540^{\circ} - 60^{\circ} \\[br] \text{or, } 60^{\circ} =2 \alpha & \text{or, }6 \alpha = 120^{\circ} & \text{or, }6 \alpha = 480^{\circ} \\[br] \text{or, } \alpha = \frac{60^{\circ}}{2} & \text{or, } \alpha = \frac{120^{\circ}}{6} & \text{or, } \alpha = \frac{480^{\circ}}{6} \\[br] \text{or, } \alpha = 30^{\circ} & \text{or, } \alpha = 20^{\circ} & \text{or, } \alpha = 80^{\circ} \\[br]\hline[br]\end{tabular} [/math][br][math] \therefore \alpha = 20^{\circ}, 30^{\circ}, 80^{\circ} [/math] [br][br]11. (a) If [math] 2\sin x \sin y = \frac{\sqrt{3}}{2}, \ \ \ \frac{1}{\tan x} + \frac{1}{\tan y } = 2 [/math] , find the minimum value of [math] x + y [/math] .[br]Solution:[br][math] [br]\begin{align} [br]\text{Given} \ & \\[br]\ & 2\sin x \sin y = \frac{\sqrt{3}}{2} ......(i) \\[br]\text{Now} \ & \\[br]\ & \frac{1}{\tan x} + \frac{1}{\tan y } = 2 \\[br]\text{or, } & \cot x + \cot y = 2 \\[br]\text{or, } & \frac{\cos x}{\sin x} + \frac{ \cos y }{\sin y} = 2 \\[br]\text{or, } & \frac{ \cos x \sin y +\sin x \cos y }{ \sin x \sin y } = 2 \\[br]\text{or, } & \sin x \cos y + \cos x \sin y = 2 \sin x \sin y \\[br]\text{or, } & \sin (x+y) = \frac{\sqrt{3}}{2}\\[br]\text{or, } & \sin (x+y) = \sin 60^{\circ} \\[br]\therefore & \ \ x+y = 60^{\circ}[br]\end{align} [br][/math][br][br](b) [math] 2\cos x \sin y = \frac{1}{\sqrt{2}} \ \ ( x> y ), [/math] [br][math] \tan x + \cot y = 2 [/math] , find [math] x-y [/math] . [br][math] [ 0^{\circ} \le (x-y) \le 360^{\circ} ] [/math] [br]Solution:[br][math] [br]\begin{align} [br]\text{Given} \ & \\[br] \ & 2\cos x \sin y = \frac{1}{\sqrt{2}} \ \ ( x> y ), ......(i) \\[br] \text{Now} \ & \\[br]\ & \tan x + \cot y = 2 \\[br]\text{or, } & \frac{\sin x}{\cos x} + \frac{ \cos y }{\sin y} = 2 \\[br]\text{or, } & \frac{ \sin x \sin y +\cos x \cos y }{ \cos x \sin y } = 2 \\[br]\text{or, } & \cos x \cos y + \sin x \sin y = 2 \cos x \sin y \\[br]\text{or, } & \cos (x-y) = \frac{1}{\sqrt{2}}\\[br]\text{or, } & \cos (x-y) = \cos 45^{\circ} , \cos (360^{\circ} - 45^{\circ}) \\[br]\therefore & \ \ x- y = 45^{\circ}, 315^{\circ} \\[br]\end{align} [br][/math][br]