8.280. $\sin 2 x \sin 6 x \cos 4 x+\frac{1}{4} \cos 12 x=0$.
## Solution.

From the condition we have

$$
\begin{aligned}
& (2 \sin 2 x \sin 6 x) \cos 4 x+\cos 12 x=0 \Leftrightarrow \\
& \Leftrightarrow 2(\cos 4 x-\cos 8 x) \cos 4 x+\cos 12 x=0 \Leftrightarrow \\
& \Leftrightarrow 2 \cos ^{2} 4 x-2 \cos 8 x \cos 4 x+\cos 12 x=0 \Leftrightarrow \\
& \Leftrightarrow 2 \cos ^{2} 4 x-\cos 4 x-\cos 12 x+\cos 12 x=0 \Leftrightarrow \\
& \Leftrightarrow 2 \cos ^{2} 4 x-\cos 4 x=0 \Leftrightarrow \cos 4 x(2 \cos 4 x-1)=0
\end{aligned}
$$

From this,

1) $\cos 4 x=0, 4 x=\frac{\pi}{2}+\pi k, x_{1}=\frac{\pi}{8}+\frac{\pi k}{4}=\frac{\pi}{8}(2 k+1), k \in Z$;
2) $2 \cos 4 x-1=0, \cos 4 x=\frac{1}{2}, 4 x= \pm \frac{\pi}{3}+2 \pi n$,

$$
x_{2}= \pm \frac{\pi}{12}+\frac{\pi n}{2}=\frac{\pi}{12}(6 n \pm 1), n \in Z
$$

Answer: $x_{1}=\frac{\pi}{8}(2 k+1) ; x_{2}=\frac{\pi}{12}(6 n \pm 1)$, where $k$ and $n \in Z$.