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