8.141. $\cos (x+1) \sin 2(x+1)=\cos 3(x+1) \sin 4(x+1)$
## Solution.

Rewrite the equation as

$$
\begin{aligned}
& \frac{1}{2}(\sin (2 x+2-x-1)+\sin (2 x+2+x+1))- \\
& -\frac{1}{2}(\sin (4 x+4-3 x-3)+\sin (4 x+4+3 x+3))=0, \quad \sin (x+1)+\sin (x+3)- \\
& -\sin (x+1)-\sin (7 x+7)=0, \sin (3 x+3)-\sin (7 x+7)=0 \Leftrightarrow \\
& \Leftrightarrow 2 \cos \frac{3 x+3+7 x+7}{2} \sin \frac{3 x+3-7 x-7}{2}=0, \cos (5 x+5) \sin (2 x+2)=0
\end{aligned}
$$

From this,

1) $\cos (5 x+5)=0, \quad 5 x+5=\frac{\pi}{2}+\pi k, x_{1}=-1+\frac{\pi}{10}+\frac{\pi k}{5}=\frac{\pi}{10}(2 k+1)-1$, $k \in Z$;
2) $\sin (2 x+2)=0,2 x+2=\pi n, \quad x_{2}=-1+\frac{\pi n}{2}, n \in Z$.

Answer: $\quad x_{1}=-1+\frac{\pi}{10}(2 k+1) x_{2}=-1+\frac{\pi n}{2}, k, n \in Z$.