$8.425 \sqrt{\cos 2 x}+\sqrt{1+\sin 2 x}=2 \sqrt{\sin x+\cos x}$.
Solution. The original equation is equivalent to the following:

$\sqrt{(\cos x-\sin x)(\cos x+\sin x)}+\sqrt{(\sin x+\cos x)^{2}}=2 \sqrt{\sin x+\cos x}$.

1) $\sin x+\cos x=0 \Leftrightarrow x=-\frac{\pi}{4}+\pi k$.
2) $\left\{\begin{array}{l}\sin x+\cos x>0 \\ \sqrt{\cos x-\sin x}+\sqrt{\cos x+\sin x}=2 .\end{array}\right.$

Square the last equation:

$\left\{\begin{array}{l}\sin x+\cos x>0 \\ \cos x-\sin x \geq 0 \\ \cos x+\sqrt{\cos 2 x}=2 .\end{array}\right.$

Since $\cos x \leq 1, \sqrt{\cos 2 x} \leq 1$, this system is equivalent to $\left\{\begin{array}{l}\cos x=1, \\ \cos 2 x=1\end{array} \Leftrightarrow x=2 \pi k\right.$

Answer: $x_{1}=\frac{\pi}{4}(4 k-1), x_{2}=2 \pi k$.