Answer
$\frac{\pi}{6}+2k\pi$, $\frac{2\pi}{3}+2k\pi$, $\frac{5\pi}{6}+2k\pi$, $\frac{4\pi}{3}+2k\pi$
Work Step by Step
$4\sin \theta\cos \theta+2\sin\theta-2\cos \theta-1=0$
$2\sin\theta(2\cos \theta+1)-(2\cos\theta+1)=0$
$(2\sin\theta-1)(2\cos\theta+1)=0$
If $2\sin\theta-1=0$, then $2\sin\theta=1$, and $\sin\theta=\frac{1}{2}$. The solutions are $\frac{\pi}{6}+2k\pi$ and $\frac{5\pi}{6}+2k\pi$.
If $2\cos\theta+1=0$, then $2\cos\theta=-1$, and $\cos\theta=-\frac{1}{2}$. The solutions are $\frac{2\pi}{3}+2k\pi$ and $\frac{4\pi}{3}+2k\pi$.
