Answer
$\frac{\pi}{2}+2k\pi$, $\frac{7\pi}{6}+2k\pi$, $\frac{11\pi}{6}+2k\pi$
Work Step by Step
$2\cos^2 \theta+\sin \theta=1$
$2(1-\sin^2 \theta)+\sin \theta=1$
$2-2\sin^2 \theta+\sin \theta=1$
$2\sin^2\theta-\sin \theta-1=0$
$(2\sin \theta+1)(\sin \theta-1)=0$
If $2\sin \theta+1=0$, then $2\sin \theta=-1$, and $\sin \theta=-\frac{1}{2}$. Then $\theta =\frac{7\pi}{6}+2k\pi$ or $\theta =\frac{11\pi}{6}+2k\pi$.
If $\sin \theta-1=0$, then $\sin \theta=1$. Then $\theta=\frac{\pi}{2}+2k\pi$.
