Answer
$\frac{\pi}{6}$, $\frac{5\pi}{6}$, $\frac{3\pi}{2}$
Work Step by Step
$\sin x=\cos 2x$
$\sin x=1-2\sin^2x$
$2\sin^2x+\sin x-1=0$
$(2\sin x-1)(\sin x+1)=0$
If $2\sin x-1=0$, then $2\sin x=1$, and $\sin x=\frac{1}{2}$. The only solutions in $[0, 2\pi)$ are $\frac{\pi}{6}$ and $\frac{5\pi}{6}$.
If $\sin x+1=0$, then $\sin x=-1$. The only solution in $[0, 2\pi)$ is $\frac{3\pi}{2}$.
