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