Answer
$\pi/3+2k\pi$, $5\pi/3+2k\pi$, $(2k+1)\pi$
Work Step by Step
$2\sin^2\theta-\cos \theta=1$
$2(1-\cos^2\theta)-\cos \theta=1$
$2-2\cos^2\theta-\cos \theta=1$
$2\cos^2\theta+\cos \theta-1=0$
$(2\cos\theta-1)(\cos \theta+1)=0$
If $2\cos \theta-1=0$, then $2\cos \theta=1$, and $\cos \theta=1/2$. Then $\theta=\pi/3+2k\pi$ or $\theta=5\pi/3+2k\pi$.
If $\cos \theta+1=0$, then $\cos \theta=-1$, and $\theta=\pi+2k\pi=(2k+1)\pi$.
