8.360. $\sqrt{2}\left(\cos ^{4} 2 x-\sin ^{4} 2 x\right)=\cos 2 x+\sin 2 x$.
## Solution.

Let's write the equation as

$$
\begin{aligned}
& \sqrt{2}\left(\cos ^{2} 2 x+\sin ^{2} 2 x\right)\left(\cos ^{2} 2 x-\sin ^{2} 2 x\right)-(\cos 2 x+\sin 2 x)=0 \Leftrightarrow \\
& \Leftrightarrow \sqrt{2}(\cos 2 x+\sin 2 x)(\cos 2 x-\sin 2 x)-(\cos 2 x+\sin 2 x)=0 \Leftrightarrow \\
& \Leftrightarrow(\cos 2 x+\sin 2 x)(\sqrt{2} \cos 2 x-\sqrt{2} \sin 2 x-1)=0
\end{aligned}
$$

From this, $\cos 2 x+\sin 2 x=0$ or $\sqrt{2} \cos 2 x-\sqrt{2} \sin 2 x-1=0$.

1) $\operatorname{tg} 2 x=-1, 2 x=-\frac{\pi}{4}+\pi k, x_{1}=-\frac{\pi}{8}+\frac{\pi k}{2}=\frac{\pi}{8}(4 k-1), k \in \mathbb{Z}$.
2) $\frac{\sqrt{2}}{2} \cos 2 x-\frac{\sqrt{2}}{2} \sin 2 x=\frac{1}{2} \Leftrightarrow \cos \frac{\pi}{4} \cos 2 x-\sin \frac{\pi}{4} \sin 2 x=\frac{1}{2} \Leftrightarrow$

$$
\Leftrightarrow \cos \left(\frac{\pi}{4}+2 x\right)=\frac{1}{2}, \frac{\pi}{4}+2 x= \pm \frac{\pi}{3}+2 \pi n, x_{2}= \pm \frac{\pi}{6}-\frac{\pi}{8}+\pi n, n \in \mathbb{Z}
$$

Answer: $\quad x_{1}=-\frac{\pi}{8}(4 k-1), k \in \mathbb{Z} ; x_{2}= \pm \frac{\pi}{6}-\frac{\pi}{8}+\pi n, n \in \mathbb{Z}$.