8.223. $\frac{1}{2} \sin 4 x \sin x+\sin 2 x \sin x=2 \cos ^{2} x$.
## Solution.

Let's write this equation as

$\sin 2 x \cos 2 x \sin x+\sin 2 x \sin x=2 \cos ^{2} x \Leftrightarrow$

$$
\begin{aligned}
& \Leftrightarrow 2 \sin x \cos x\left(2 \cos ^{2} x-1\right) \sin x+2 \sin x \cos x \sin x=2 \cos ^{2} x \Leftrightarrow \\
& \Leftrightarrow 2 \cos x\left(2 \cos ^{2} x-1\right) \sin ^{2} x+2 \sin ^{2} x \cos x-2 \cos ^{2} x=0 \Leftrightarrow \\
& \Leftrightarrow 2 \cos x\left(\left(2 \cos ^{2} x-1\right)\left(1-\cos ^{2} x\right)+1-\cos ^{2} x-\cos x\right)=0 \Leftrightarrow \\
& \Leftrightarrow \cos x\left(2 \cos ^{2} x-2 \cos ^{4} x-\cos x\right)=0 \Leftrightarrow \\
& \Leftrightarrow-\cos ^{2} x\left(2 \cos ^{3} x-2 \cos x+1\right)=0
\end{aligned}
$$

From this, we have $\cos ^{2} x=0, \cos x=0, x=\frac{\pi}{2}+\pi k=\frac{\pi}{2}(2 k+1), k \in Z$; $2 \cos ^{3} x-2 \cos x+1 \neq 0$.

Answer: $x=\frac{\pi}{2}(2 k+1), k \in Z$.