Answer
$(\frac{1}{2},\frac{\pi}{6})$ and $(\frac{1}{2},\frac{5\pi}{6})$
Work Step by Step
$r=\sin \theta$ and $r=1-\sin\theta$ (Add both equations)
$2r=1$
$r=\frac{1}{2}$
Find $\theta\in [0,2\pi]$ such that $r=\frac{1}{2}$:
$r=\sin \theta$
$\frac{1}{2}=\sin\theta$
$\theta=\frac{\pi}{6}\vee \theta=\frac{5\pi}{6}$
Thus, the intersection points are $(\frac{1}{2},\frac{\pi}{6})$ and $(\frac{1}{2},\frac{5\pi}{6})$.
