8.211. $\cos x \cos 2 x \sin 3 x=0.25 \sin 2 x$.
## Solution.

Let's write the equation as:

$\cos x \cos 2 x \sin 3 x - 0.5 \sin x \cos x = 0 \Leftrightarrow \cos x(\cos 2 x \sin 3 x - 0.5 \sin x) = 0$.

## From this, either

$\cos x = 0, x_{1} = \frac{\pi}{2} + \pi k = \frac{\pi}{2}(2 k + 1), k \in \mathbb{Z}$,

## or

$\cos 2 x \sin 3 x - 0.5 \sin x = 0$, or $2 \cos 2 x \sin 3 x - \sin x = 0$.

Using the formula $\sin \alpha \cos \beta = \frac{1}{2}(\sin (\alpha - \beta) + \sin (\alpha + \beta))$, we have:

$\sin (3 x - 2 x) + \sin (3 x + 2 x) - \sin x = 0 \Leftrightarrow \sin x + \sin 5 x - \sin x = 0 \Leftrightarrow$

$\Leftrightarrow \sin 5 x, 5 x = \pi n, x_{2} = \frac{\pi n}{5}, n \in \mathbb{Z}$.

Answer: $x_{1} = \frac{\pi}{2}(2 k + 1), x_{2} = \frac{\pi n}{5}$, where $k$ and $n \in \mathbb{Z}$.