8.103. $2 \cos 2 x+2 \tan^{2} x=5$.
## Solution.

Domain of definition: $\cos x \neq 0$.

Rewrite the equation as

$$
\begin{aligned}
& 2 \cos 2 x+\frac{2(1-\cos 2 x)}{1+\cos 2 x}-5=0 \Rightarrow 2 \cos 2 x(1+\cos 2 x)+2(1-\cos 2 x)- \\
& -5(1+\cos 2 x)=0,2 \cos ^{2} 2 x-5 \cos 2 x-3=0 .
\end{aligned}
$$

Solving this equation as a quadratic in $\cos 2 x$, we find $\cos 2 x=3, \varnothing$, or $\cos 2 x=-\frac{1}{2}, 2 x= \pm \frac{2}{3} \pi+2 \pi k$, $x= \pm \frac{\pi}{3}+\pi k=\frac{\pi}{3}(3 k \pm 1), k \in Z$.

Answer: $\quad x=\frac{\pi}{3}(3 k \pm 1), k \in Z$.