8.202. $2 \cos 13 x+3 \cos 3 x+3 \cos 5 x-8 \cos x \cos ^{3} 4 x=0$.
## Solution.

Since

$$
\begin{aligned}
& \cos 3 \alpha=4 \cos ^{3} \alpha-3 \cos \alpha, \cos \alpha+\cos \beta=2 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2} \\
& \cos \alpha-\cos \beta=2 \sin \frac{\alpha+\beta}{2} \sin \frac{\beta-\alpha}{2}
\end{aligned}
$$

the equation can be written as

$2 \cos 13 x+3(\cos 3 x+\cos 5 x)-8 \cos x \cos ^{3} 4 x=0 \Leftrightarrow$
$\Leftrightarrow 2 \cos 13 x+3 \cdot 2 \cos 4 x \cos x-8 \cos x \cos ^{3} 4 x=0 \Leftrightarrow$

$\Leftrightarrow 2 \cos 13 x-2 \cos x\left(4 \cos ^{3} 4 x-3 \cos 4 x\right)=0 \Leftrightarrow$

$\Leftrightarrow 2 \cos 13 x-2 \cos x \cos 12 x=0 \Leftrightarrow 2 \cos 13 x-(\cos 13 x+\cos 11 x)=0 \Leftrightarrow$ $\Leftrightarrow-2 \sin 12 x \sin x=0$.

From this,

1) $\sin 12 x=0, 12 x=\pi k, x_{1}=\frac{\pi k}{12}, k \in \mathbb{Z}$;
2) $\sin x=0, x_{2}=\pi n, n \in \mathbb{Z}; x_{2}$ is included in $x_{1}$.

Answer: $x=\frac{\pi k}{12}, k \in \mathbb{Z}$.