8.331. $\cos 8 x+3 \cos 4 x+3 \cos 2 x=8 \cos x \cos ^{3} 3 x-0.5$.
## Solution.

Let's rewrite the equation as

$$
\begin{aligned}
& \cos 8 x + 3 \cos 4 x + 3 \cos 2 x = 2(2 \cos x \cos 3 x)\left(2 \cos ^{2} 3 x\right) - 0.5 \Leftrightarrow \\
& \Leftrightarrow \cos 8 x + 3 \cos 4 x + 3 \cos 2 x = 2(\cos 2 x + \cos 4 x)(1 + \cos 6 x) - 0.5 \Leftrightarrow \\
& \Leftrightarrow \cos 8 x + 3 \cos 4 x + 3 \cos 2 x = 2 \cos 2 x + 2 \cos 4 x + 2 \cos 2 x \cos 6 x + \\
& + 2 \cos 4 x \cos 6 x - 0.5 \Leftrightarrow \\
& \Leftrightarrow \cos 8 x + \cos 4 x + \cos 2 x = \cos 4 x + \cos 8 x + \cos 2 x + \cos 10 x - 0.5 \\
& \cos 10 x = \frac{1}{2}, 10 x = \pm \frac{\pi}{3} + 2 \pi k, x = \pm \frac{\pi}{30} + \frac{\pi k}{5} = \frac{\pi}{30}(6 k \pm 1), k \in \mathbb{Z}
\end{aligned}
$$

Answer: $\frac{\pi}{30}(6 k \pm 1), k \in \mathbb{Z}$.