The general solution of  $2^{\sin x}+2^{\cos x}=2^{1-\frac{1}{\sqrt{2}}}$ is

$\frac{2^{\sin x}+2^{\cos x}}{2}\ge \sqrt{2^{\sin x}\cdot 2^{\cos x}}$

$2^{\sin x}+2^{\cos x}\ge 2\cdot \sqrt{2^{\sin x+\cos x}}$

But minimum value of  $\sin x+\cos x$ is  $-\sqrt{2}$

$\therefore 2^{\sin x}+2^{\cos x}\ge 2\cdot \sqrt{2^{-\sqrt{2}}}=2^{1-\frac{1}{\sqrt{2}}}$

The equation holds when  $2^{\sin x}=2^{\cos x}$

$\Rightarrow \sin x=\cos x$  or  $\tan x=1$

$\Rightarrow x=n\pi +\frac{\pi }{4}$.