8.107. $4 \sin x + \cos x = 4$.
Solution.

Rewrite the equation as

$$
\begin{aligned}
& 4 \sin 2\left(\frac{x}{2}\right)+\cos 2\left(\frac{x}{2}\right)-4\left(\cos ^{2} \frac{x}{2}+\sin ^{2} \frac{x}{2}\right)=0 \Leftrightarrow 8 \sin \frac{x}{2} \cos \frac{x}{2}+ \\
& +\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}-4 \cos ^{2} \frac{x}{2}-4 \sin ^{2} \frac{x}{2}=0,5 \sin ^{2} \frac{x}{2}-8 \sin \frac{x}{2} \cos \frac{x}{2}+ \\
& +3 \cos ^{2} \frac{x}{2}=0, \Leftrightarrow 5 \operatorname{tg}^{2} \frac{x}{2}-8 \operatorname{tg} \frac{x}{2}+3=0
\end{aligned}
$$

Solving the last equation as a quadratic equation in terms of $\operatorname{tg} \frac{x}{2}$, we find

$$
\begin{aligned}
& \left(\operatorname{tg} \frac{x}{2}\right)_{1}=\frac{3}{5},\left(\frac{x}{2}\right)_{1}=\operatorname{arctg} \frac{3}{5}+\pi n, x_{1}=2 \operatorname{arctg} \frac{3}{5}+2 \pi n, n \in Z ;\left(\operatorname{tg} \frac{x}{2}\right)_{2}=1 \\
& \left(\frac{x}{2}\right)_{2}=\frac{\pi}{4}+\pi k, x_{2}=\frac{\pi}{2}+2 \pi k=\frac{\pi}{2}(4 k+1), k \in Z
\end{aligned}
$$

Answer: $\quad x_{1}=2 \operatorname{arctg} \frac{3}{5}+2 \pi n ; x_{2}=\frac{\pi}{2}(4 k+1), \quad n, k \in Z$