If $2 {\sin }^2\left(\frac{\mathrm{\pi }}{2}{\cos }^2\mathrm{x}\right)=1-\cos \left(\mathrm{\pi } \sin 2\mathrm{x}\right), \mathrm{x}\ne \left(2\mathrm{n}+1\right)\frac{\mathrm{\pi }}{2}, \mathrm{n}\in \mathrm{I}$ then cos2x is equal to

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

$\Rightarrow \cos \mathrm{x}\left(\cos \mathrm{x}-2 \mathrm{sinx}\right)=0\cos \mathrm{x}\ne 0\mathrm{x}\ne \left(2\mathrm{n}+1\right)\frac{\overline{\mathrm{n}}}{2},\tan \mathrm{x}=\frac{1}{2}$$\Rightarrow \cos 2\mathrm{x}=\frac{1-{\tan }^2\mathrm{x}}{1+{\tan }^2\mathrm{x}}=\frac{3}{5}$