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

Explanation:

Step1: Set the two equations equal

Set $3 + 3\cos(\theta)=3 - 3\sin(\theta)$. Subtract 3 from both sides: $3\cos(\theta)=- 3\sin(\theta)$. Then $\cos(\theta)=-\sin(\theta)$. Dividing both sides by $\cos(\theta)$ (assuming $\cos(\theta)\neq0$), we get $\tan(\theta)= - 1$. The solutions of $\tan(\theta)=-1$ in the interval $[0,2\pi)$ are $\theta=\frac{3\pi}{4}$ and $\theta = \frac{7\pi}{4}$.

Step2: Find the $r$ - values

When $\theta=\frac{3\pi}{4}$: $r = 3+3\cos(\frac{3\pi}{4})=3 + 3\times(-\frac{\sqrt{2}}{2})=3-\frac{3\sqrt{2}}{2}$. When $\theta=\frac{7\pi}{4}$: $r = 3+3\cos(\frac{7\pi}{4})=3 + 3\times\frac{\sqrt{2}}{2}=3+\frac{3\sqrt{2}}{2}$. We also need to check for the case when $r = 0$. For $r = 3 + 3\cos(\theta)=0$, $\cos(\theta)=-1$, $\theta=\pi$, and $r = 3-3\sin(\pi)=3\neq0$. For $r = 3 - 3\sin(\theta)=0$, $\sin(\theta)=1$, $\theta=\frac{\pi}{2}$, and $r = 3+3\cos(\frac{\pi}{2})=3\neq0$.

Step3: Convert to rectangular - coordinates

The conversion formulas from polar to rectangular coordinates are $x = r\cos(\theta)$ and $y=r\sin(\theta)$. When $\theta=\frac{3\pi}{4}$ and $r = 3-\frac{3\sqrt{2}}{2}$: $x=(3 - \frac{3\sqrt{2}}{2})\cos(\frac{3\pi}{4})=(3 - \frac{3\sqrt{2}}{2})\times(-\frac{\sqrt{2}}{2})=-\frac{3\sqrt{2}}{2}+\frac{3}{2}$ $y=(3 - \frac{3\sqrt{2}}{2})\sin(\frac{3\pi}{4})=(3 - \frac{3\sqrt{2}}{2})\times\frac{\sqrt{2}}{2}=\frac{3\sqrt{2}}{2}-\frac{3}{2}$ When $\theta=\frac{7\pi}{4}$ and $r = 3+\frac{3\sqrt{2}}{2}$: $x=(3+\frac{3\sqrt{2}}{2})\cos(\frac{7\pi}{4})=(3+\frac{3\sqrt{2}}{2})\times\frac{\sqrt{2}}{2}=\frac{3\sqrt{2}}{2}+\frac{3}{2}$ $y=(3+\frac{3\sqrt{2}}{2})\sin(\frac{7\pi}{4})=(3+\frac{3\sqrt{2}}{2})\times(-\frac{\sqrt{2}}{2})=-\frac{3\sqrt{2}}{2}-\frac{3}{2}$

Answer:

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