Use algebra and a graph to find all points where the curves [math]r=3sin\left(\theta\right)[/math] and [math]r=3cos\left(2\theta\right)[/math] intersect.[br][br][b][u]Proof:[/u][/b][br]Proof: Consider the following polar equations: [math]r=3sin\left(\theta\right)[/math] and [math]r=3cos\left(2\theta\right)[/math]. Furthermore, note that [math]3sin\left(\theta\right)=3cos\left(2\theta\right)[/math]. Now, we shall notice the following:[br][br][math]3sin\left(\theta\right)=3cos\left(2\theta\right)[/math][br][math]\Longrightarrow sin\left(\theta\right)=cos\left(2\theta\right)[/math][br][math]\Longrightarrow sin\left(\theta\right)=1-2sin^2\left(\theta\right)[/math] -- [trig identity: [math]cos\left(2\theta\right)=1-2sin^2\left(\theta\right)[/math]][br][math]\Longrightarrow2sin^2\left(\theta\right)+sin\left(\theta\right)-1=0[/math][br][math]\Longrightarrow\left(2sin\left(\theta\right)-1\right)\left(sin\left(\theta\right)+1\right)=0[/math][br][br]Note that we can now find [math]\theta[/math] for which [math]2sin\left(\theta\right)-1=0[/math] and [math]sin\left(\theta\right)+1=0[/math]. Consider the following:[br][br]First consideration:[br][math]2sin\left(\theta\right)-1=0[/math][br][math]\Longrightarrow2sin\left(\theta\right)=1[/math][br][math]\Longrightarrow sin\left(\theta\right)=\frac{1}{2}[/math][br][math]\Longrightarrow\theta=sin^{-1}\left(\frac{1}{2}\right)[/math][br][math]\Longrightarrow\theta=\frac{\pi}{6},\frac{5\pi}{6}[/math].[br][br]Second consideration:[br][math]sin\left(\theta\right)+1=0[/math][br][math]\Longrightarrow sin\left(\theta\right)=-1[/math][br][math]\Longrightarrow\theta=sin^{-1}\left(-1\right)[/math][br][math]\Longrightarrow\theta=\frac{3\pi}{2}[/math].[br][br][br]Thus, we know that [math]\frac{\pi}{6},\frac{5\pi}{6},\frac{3\pi}{2}[/math] are all values for theta for which [math]r=3sin\left(\theta\right)[/math] and [math]r=3cos\left(2\theta\right)[/math] intersect. If we wish to represent them as points, we can substitute our values for theta into either equation and get that the points are as follows:[br][br]A = ([math]\pi/6[/math], 1.5)[br]B = ([math]5\pi/6[/math], 1.5)[br]C = ([math]3\pi/2[/math], -3).[br][br][br]Note that (0, 0) is not a point of intersection because substituting 0 in for theta yields two different values of r.