find all points of intersection between the curves.\n$r = \\sqrt{3} \\cos(3\\theta)$ and $r =…

Explanation:

Step1: Set the two equations equal

Set $\sqrt{3}\cos(3\theta)=\sin(3\theta)$. Then $\frac{\sin(3\theta)}{\cos(3\theta)}=\sqrt{3}$, so $\tan(3\theta)=\sqrt{3}$.

Step2: Solve for $3\theta$

We know that $\tan x=\sqrt{3}$ has solutions $x = \frac{\pi}{3}+k\pi$, where $k\in\mathbb{Z}$. So $3\theta=\frac{\pi}{3}+k\pi$, then $\theta=\frac{\pi}{9}+\frac{k\pi}{3}$.

Step3: Find $r$ values

When $\theta=\frac{\pi}{9}$, $r = \sin(3\times\frac{\pi}{9})=\sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}$. When $\theta=\frac{\pi}{9}+\frac{\pi}{3}=\frac{4\pi}{9}$, $r=\sin(3\times\frac{4\pi}{9})=\sin(\frac{4\pi}{3})=-\frac{\sqrt{3}}{2}$. When $\theta=\frac{\pi}{9}+\frac{2\pi}{3}=\frac{7\pi}{9}$, $r=\sin(3\times\frac{7\pi}{9})=\sin(\frac{7\pi}{3})=\frac{\sqrt{3}}{2}$. In polar coordinates, $(r,\theta)$ and $(-r,\theta + \pi)$ represent the same point. The points of intersection in polar - coordinates are $(\frac{\sqrt{3}}{2},\frac{\pi}{9}),(\frac{\sqrt{3}}{2},\frac{7\pi}{9})$.

Answer:

$(\frac{\sqrt{3}}{2},\frac{\pi}{9}),(\frac{\sqrt{3}}{2},\frac{7\pi}{9})$