If $3\cos x\ne 2\sin x$ , then the general solution of ${\sin }^{2}x-\cos 2x=2-\sin 2x$  is

${\sin }^{2}x-\cos 2x=2-\sin 2x$

$\frac{1-\cos 2x}{2}-\cos 2x=2-\sin 2x$

$\therefore 3\cos 2x-2\sin 2x=-3$

$\Rightarrow 2\sin 2x=3\left(1+\cos 2x\right)$

$\Rightarrow 4\sin x\cos x=6{\cos }^{2}x$

$\Rightarrow \cos x\left(4\sin x-6\cos x\right)=0$

$\cos x=0\left(or\right)4\sin x-6\cos x=0$

$4\sin x-6\cos x=0$  is not possible  by given condition

$\therefore \cos x=0$

$x=\left(4n\pm 1\right)\frac{\pi }{2}:n\in z$