$2{\sin }^2\left(\frac{\pi }{2}{\cos }^{2}x\right)=1-\cos (\pi \sin 2x)$ is satisfied  by

$2{\sin }^2\left(\frac{\pi }{2}{\cos }^{2}x\right)=1-\cos (\pi \sin 2x)$

$\Rightarrow 2{\sin }^2\left\{\left(\frac{\pi }{2}\right){\cos }^{2}x\right\}=2{\sin }^2\left\{\left(\frac{\pi }{2}\right)\sin 2x\right\}$

$\Rightarrow \left(\frac{\pi }{2}\right){\cos }^{2}x=\pm \left(\frac{\pi }{2}\right)\sin 2x$

$\Rightarrow \cos x=0\text{ or }\cos x=\pm 2\sin x$$\Rightarrow x=(2n+1)\frac{\pi }{2},n\in Z\text{ or }\tan x=\pm \frac{1}{2}$