Answer
The two points of intersection are: $(\frac{4+\sqrt{2}}{2}, \frac{\pi}{4})$ and $(\frac{4-\sqrt{2}}{2}, \frac{5\pi}{4})$
Work Step by Step
$r = 2+sin~\theta$
$r = 2+cos~\theta$
To find the points of intersection, we can equate the expressions for $r$:
$2+sin~\theta = 2+cos~\theta$
$sin~\theta = cos~\theta$
$\frac{sin~\theta}{cos~\theta} = 1$
$tan~\theta = 1$
$\theta = \frac{\pi}{4}, \frac{5\pi}{4}$
We can find $r$ when $\theta = \frac{\pi}{4}$:
$r = 2+sin~\theta$
$r = 2+sin~\frac{\pi}{4}$
$r = 2+\frac{\sqrt{2}}{2}$
$r = \frac{4+\sqrt{2}}{2}$
We can find $r$ when $\theta = \frac{5\pi}{4}$:
$r = 2+sin~\theta$
$r = 2+sin~\frac{5\pi}{4}$
$r = 2-\frac{\sqrt{2}}{2}$
$r = \frac{4-\sqrt{2}}{2}$
The two points of intersection are: $(\frac{4+\sqrt{2}}{2}, \frac{\pi}{4})$ and $(\frac{4-\sqrt{2}}{2}, \frac{5\pi}{4})$
