Answer
$\frac{\pi}{2}$, $\frac{7\pi}{6}$, $\frac{3\pi}{2}$, $\frac{11\pi}{6}$
Work Step by Step
$\sin 2\theta+\cos \theta=0$
$2\sin \theta\cos \theta + \cos \theta =0$
$\cos \theta(2\sin \theta+1)=0$
$\cos \theta=0$ or $2\sin \theta+1=0$
If $\cos \theta=0$, $\theta=\frac{\pi}{2}$ or $\theta=\frac{3\pi}{2}$.
If $2\sin \theta+1=0$, then $2\sin \theta=-1$, so $\sin \theta=-\frac{1}{2}$, so $\theta=\frac{7\pi}{6}$ or $\theta=\frac{11\pi}{6}$.
So the only solutions in the interval $[0, 2\pi)$ are $\theta=\frac{\pi}{2}$, $\theta=\frac{7\pi}{6}$, $\theta=\frac{3\pi}{2}$, and $\theta=\frac{11\pi}{6}$.
