Solve the following equations:

$2^{\sin ^{2} x}+2^{\cos ^{2} x}=2 \sqrt{2}$

$\Rightarrow 2^{\sin ^{2} x}+2^{1-\sin ^{2} x}=2 \sqrt{2}$

$\Rightarrow 2^{\sin ^{2} x}+\frac{2}{2^{\sin ^{2} x}}=2 \sqrt{2}$

Let $2^{\sin ^{2} x}=y$

$\Rightarrow y+\frac{2}{y}=2 \sqrt{2}$

$\Rightarrow y^{2}+2=2 \sqrt{2} y$

$\Rightarrow y^{2}-2 \sqrt{2} y+2=0$

$\Rightarrow y^{2}-\sqrt{2} y-\sqrt{2} y+2=0$

$\Rightarrow y(y-\sqrt{2})-\sqrt{2}(y-\sqrt{2})=0$

$\Rightarrow(y-\sqrt{2})^{2}=0$

$\Rightarrow(y-\sqrt{2})=0$

$\Rightarrow y=\sqrt{2}$

$\Rightarrow 2^{\sin ^{2} x}=2^{\frac{1}{2}}$

$\Rightarrow \sin ^{2} x=\frac{1}{2}$

$\Rightarrow \sin ^{2} x=\sin ^{2} \frac{\pi}{4}$

$\Rightarrow x=n \pi \pm \frac{\pi}{4}, n \in \mathbb{Z}$