The number of solutions of  $(\sin 2\mathrm{x}+\cos 2\mathrm{x}{)}^{1+\sin 4\mathrm{x}}=2\text{ in }$$[-\mathrm{\pi },\mathrm{\pi }]$ is equal to

$\begin{array}{l}(\sin 2\mathrm{x}+\cos 2\mathrm{x}{)}^{(\sin 2\mathrm{x}+\cos 2\mathrm{x}{)}^2}=2\\ \Rightarrow \sin 2\mathrm{x}+\cos 2\mathrm{x}=\sqrt{2}\\ \Rightarrow \sin \left(2\mathrm{x}+\frac{\mathrm{\pi }}{4}\right)=1\\ \Rightarrow 2\mathrm{x}+\frac{\mathrm{\pi }}{4}=2\mathrm{n\pi }+\frac{\mathrm{\pi }}{2}\Rightarrow \mathrm{x}=\mathrm{n\pi }+\frac{\mathrm{\pi }}{8}\end{array}$

for, $\mathrm{x}\in [-\mathrm{\pi },\mathrm{\pi }],\text{ put }\mathrm{n}=-1,0,1$

we get, $\mathrm{x}=-\frac{7\mathrm{\pi }}{8},\frac{\mathrm{\pi }}{8}$