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

From the condition

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

Solving this equation as a quadratic in $\sin (2 x+1)$, we find

$$
(\sin (2 x+1))_{1}=\frac{1}{2}, 2 x_{1}+1=(-1)^{k} \frac{\pi}{6}+\pi k, x_{1}=(-1)^{k} \frac{\pi}{12}-\frac{1}{2}+\frac{\pi k}{2}, k \in Z
$$

$$
(\sin (2 x+1))_{2}=1,2 x_{2}+1=\frac{\pi}{2}+2 \pi n, x_{2}=\frac{\pi}{4}-\frac{1}{2}+\pi n=\frac{\pi}{4}(4 n+1)-\frac{1}{2}, n \in Z
$$

Answer: $\quad x_{1}=(-1)^{k} \frac{\pi}{12}-\frac{1}{2}+\frac{\pi k}{2} ; x_{2}=\frac{\pi}{4}(4 n+1)-\frac{1}{2}, k, n \in Z$