Answer
$\pi/6+2k\pi$, $5\pi/6+2k\pi$
Work Step by Step
$\cos 2\theta=3\sin \theta-1$
$1-2\sin^2\theta=3\sin \theta-1$
$2\sin^2\theta+3\sin \theta-2=0$
$(2\sin \theta-1)(\sin \theta+2)=0$
If $2\sin \theta-1=0$, then $2\sin \theta=1$, and $\sin \theta=1/2$. Then $\theta=\pi/6+2k\pi$ or $\theta=5\pi/6+2k\pi$.
If $\sin \theta+2=0$, then $\sin \theta=-2$, which is impossible because $\sin \theta$ cannot be less than -1.
So $\pi/6+2k\pi$, and $5\pi/6+2k\pi$ are the only solutions.
