find all points of intersection between the curves. $r = \\cos(3\\theta)$ and $r = \\sin(3\\theta)$. enter…

Explanation:

Step1: Set the two equations equal

Set $\cos(3\theta)=\sin(3\theta)$. Then $\tan(3\theta) = 1$.

Step2: Solve for $3\theta$

We know that if $\tan x=1$, then $x = n\pi+\frac{\pi}{4}$, where $n\in\mathbb{Z}$. So $3\theta=n\pi+\frac{\pi}{4}$, and $\theta=\frac{n\pi}{3}+\frac{\pi}{12}$.

Step3: Find values of $\theta$ in the range $[0, 2\pi)$

When $n = 0$, $\theta=\frac{\pi}{12}$; when $n = 1$, $\theta=\frac{\pi}{3}+\frac{\pi}{12}=\frac{5\pi}{12}$; when $n = 2$, $\theta=\frac{2\pi}{3}+\frac{\pi}{12}=\frac{3\pi}{4}$; when $n = 3$, $\theta=\pi+\frac{\pi}{12}=\frac{13\pi}{12}$; when $n = 4$, $\theta=\frac{4\pi}{3}+\frac{\pi}{12}=\frac{17\pi}{12}$; when $n = 5$, $\theta=\frac{5\pi}{3}+\frac{\pi}{12}=\frac{7\pi}{4}$.

Step4: Find the $r$ - values

Substitute $\theta$ into either $r = \cos(3\theta)$ or $r=\sin(3\theta)$. When $\theta=\frac{\pi}{12}$, $r=\cos(\frac{\pi}{4})=\frac{\sqrt{2}}{2}$; when $\theta=\frac{5\pi}{12}$, $r=\cos(\frac{5\pi}{4})=-\frac{\sqrt{2}}{2}$; when $\theta=\frac{3\pi}{4}$, $r=\cos(\frac{9\pi}{4})=\frac{\sqrt{2}}{2}$; when $\theta=\frac{13\pi}{12}$, $r=\cos(\frac{13\pi}{4})=-\frac{\sqrt{2}}{2}$; when $\theta=\frac{17\pi}{12}$, $r=\cos(\frac{17\pi}{4})=\frac{\sqrt{2}}{2}$; when $\theta=\frac{7\pi}{4}$, $r=\cos(\frac{21\pi}{4})=-\frac{\sqrt{2}}{2}$. The polar - coordinate points are $(\frac{\sqrt{2}}{2},\frac{\pi}{12}),(-\frac{\sqrt{2}}{2},\frac{5\pi}{12}),(\frac{\sqrt{2}}{2},\frac{3\pi}{4}),(-\frac{\sqrt{2}}{2},\frac{13\pi}{12}),(\frac{\sqrt{2}}{2},\frac{17\pi}{12}),(-\frac{\sqrt{2}}{2},\frac{7\pi}{4})$. In rectangular coordinates, for a polar point $(r,\theta)$, $x = r\cos\theta$ and $y = r\sin\theta$. For $(r,\theta)=(\frac{\sqrt{2}}{2},\frac{\pi}{12})$: $x=\frac{\sqrt{2}}{2}\cos(\frac{\pi}{12})=\frac{\sqrt{2}}{2}\times\frac{\sqrt{6}+\sqrt{2}}{4}=\frac{\sqrt{3} + 1}{4}$, $y=\frac{\sqrt{2}}{2}\sin(\frac{\pi}{12})=\frac{\sqrt{2}}{2}\times\frac{\sqrt{6}-\sqrt{2}}{4}=\frac{\sqrt{3}-1}{4}$ For $(r,\theta)=(-\frac{\sqrt{2}}{2},\frac{5\pi}{12})$: $x=-\frac{\sqrt{2}}{2}\cos(\frac{5\pi}{12})=-\frac{\sqrt{2}}{2}\times\frac{\sqrt{6}-\sqrt{2}}{4}=\frac{1 - \sqrt{3}}{4}$, $y=-\frac{\sqrt{2}}{2}\sin(\frac{5\pi}{12})=-\frac{\sqrt{2}}{2}\times\frac{\sqrt{6}+\sqrt{2}}{4}=-\frac{\sqrt{3}+1}{4}$ For $(r,\theta)=(\frac{\sqrt{2}}{2},\frac{3\pi}{4})$: $x=\frac{\sqrt{2}}{2}\cos(\frac{3\pi}{4})=-\frac{1}{2}$, $y=\frac{\sqrt{2}}{2}\sin(\frac{3\pi}{4})=\frac{1}{2}$ For $(r,\theta)=(-\frac{\sqrt{2}}{2},\frac{13\pi}{12})$: $x=-\frac{\sqrt{2}}{2}\cos(\frac{13\pi}{12})=-\frac{\sqrt{2}}{2}\times(-\frac{\sqrt{6}-\sqrt{2}}{4})=\frac{\sqrt{3}-1}{4}$, $y=-\frac{\sqrt{2}}{2}\sin(\frac{13\pi}{12})=-\frac{\sqrt{2}}{2}\times(-\frac{\sqrt{6}+\sqrt{2}}{4})=\frac{\sqrt{3}+1}{4}$ For $(r,\theta)=(\frac{\sqrt{2}}{2},\frac{17\pi}{12})$: $x=\frac{\sqrt{2}}{2}\cos(\frac{17\pi}{12})=\frac{\sqrt{2}}{2}\times(-\frac{\sqrt{6}+\sqrt{2}}{4})=-\frac{\sqrt{3}+1}{4}$, $y=\frac{\sqrt{2}}{2}\sin(\frac{17\pi}{12})=\frac{\sqrt{2}}{2}\times(-\frac{\sqrt{6}-\sqrt{2}}{4})=-\frac{\sqrt{3}-1}{4}$ For $(r,\theta)=(-\frac{\sqrt{2}}{2},\frac{7\pi}{4})$: $x=-\frac{\sqrt{2}}{2}\cos(\frac{7\pi}{4})=-\frac{1}{2}$, $y=-\frac{\sqrt{2}}{2}\sin(\frac{7\pi}{4})=\frac{1}{2}$

The intersection points in rectangular coordinates are $(\frac{\sqrt{3}+1}{4},\frac{\sqrt{3}-1}{4}),(\frac{1 - \sqrt{3}}{4},-\frac{\sqrt{3}+1}{4}),(-\frac{1}{2},\frac{1}{2}),(\frac{\sqrt{3}-1}{4},\frac{\sqrt{3}+1}{4}),(-\frac{\sqrt{3}+1}{4},-\frac{\sqrt{3}-1}{4}),(-\frac{1}{2},\frac{1}{2})$

Answer:

$(\frac{\sqrt{3}+1}{4},\frac{\sqrt{3}-1}{4}),(\frac{1 - \sqrt{3}}{4},-\frac{\sqrt{3}+1}{4}),(-\frac{1}{2},\frac{1}{2}),(\frac{\sqrt{3}-1}{4},\frac{\sqrt{3}+1}{4}),(-\frac{\sqrt{3}+1}{4},-\frac{\sqrt{3}-1}{4}),(-\frac{1}{2},\frac{1}{2})$