Answer
$(0, \frac{\pi}{2}),(\sqrt 3, \frac{\pi}{6}), (-\sqrt 3, \frac{5\pi}{6})$
Work Step by Step
$r=cot \theta$, and $r= 2cos \theta$
$cot \theta= 2cos \theta$
$cot \theta- 2cos \theta=0$
$\frac{cos \theta}{sin \theta}- 2cos \theta=0$
$sin \theta=\frac{1}{2}$ and $ \theta=\frac{\pi}{6}, \frac{5 \pi}{6}$
Now we will find the points.
$(r, \theta): (2 cos \frac{\pi}{2}, \frac{\pi}{2})=(0, \frac{\pi}{2})$
$ (2 cos \frac{\pi}{6}, \frac{\pi}{6})=(\sqrt 3, \frac{\pi}{6})$
$ (2 cos \frac{5\pi}{6}, \frac{5\pi}{6})=(-\sqrt 3, \frac{5\pi}{6})$
Hence, $(0, \frac{\pi}{2}),(\sqrt 3, \frac{\pi}{6}), (-\sqrt 3, \frac{5\pi}{6})$
