pt $\Leftrightarrow 2\sin x+\sin2x-\cos x-\cos^2x=\sin^2x$$\Leftrightarrow 2\sin x(1+\cos x)-\cos x-1=0$
$\Leftrightarrow (1+\cos x)(2\sin x-1)=0$
$\Leftrightarrow \cos =-1 or \sin x=\frac{1}{2}$
$\Leftrightarrow x=\frac{-\pi}{2}+k2\pi; x=\frac{\pi}{6}+k2\pi; x=\frac{5\pi}{6}+k2\pi, k\in Z$