Answer
$\frac{\pi}{4}$, $\frac{3\pi}{4}$, $\frac{5\pi}{4}$, $\frac{7\pi}{4}$
Work Step by Step
$\cos 2x\csc^2 x=2\cos 2x$
$\csc^2x=2$
$\sin^2x=\frac{1}{2}$
$\sin x=\pm\frac{\sqrt{2}}{2}$
If $\sin x=\frac{\sqrt{2}}{2}$, then the only solutions in $[0, 2\pi)$ are $\frac{\pi}{4}$ and $\frac{3\pi}{4}$.
If $\sin x=-\frac{\sqrt{2}}{2}$, then the only solutions in $[0, 2\pi)$ are $\frac{5\pi}{4}$ and $\frac{7\pi}{4}$.
