8.297. $4 \sin 5 x \cos 5 x\left(\cos ^{4} x-\sin ^{4} x\right)=\sin 4 x$.

8.297. $4 \sin 5 x \cos 5 x\left(\cos ^{4} x-\sin ^{4} x\right)=\sin 4 x$.
## Solution.

Let's write this equation in the form

$2 \sin 10 x\left(\cos ^{2} x+\sin ^{2} x\right)\left(\cos ^{2} x-\sin ^{2} x\right)-2 \sin 2 x \cos 2 x=0 \Leftrightarrow$

$\Leftrightarrow 2 \sin 10 x \cos 2 x-2 \sin 2 x \cos 2 x=0, 2 \cos 2 x(\sin 10 x-\sin 2 x)=0$.

From this,

1) $\cos 2 x=0, 2 x=\frac{\pi}{2}+\pi k, x_{1}=\frac{\pi}{4}+\frac{\pi k}{2}, k \in Z$;
2) $\sin 10 x-\sin 2 x=0 \Leftrightarrow 2 \cos 6 x \sin 4 x=0$.

From this,

1) $\cos 6 x=0, 6 x=\frac{\pi}{2}+\pi n, x_{2}=\frac{\pi}{12}+\frac{\pi n}{6}=\frac{\pi}{12}(2 n+1), n \in Z$;
2) $\sin 4 x=0, 4 x=\pi l, x_{3}=\frac{\pi l}{4}, l \in Z ; x_{1}$ is included in $x_{2}$.

Answer: $x_{1}=\frac{\pi k}{4} ; x_{2}=\frac{\pi}{12}(2 n+1)$, where $k$ and $n \in Z$.