8.293. $\frac{4 \sin \left(\frac{\pi}{6}+x\right) \sin \left(\frac{5 \pi}{6}+x\right)}{\cos ^{2} x}+2 \tan x=0$.
## Solution.

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

Write the equation in the form

$$
\begin{aligned}
& \frac{2\left(\cos \left(\frac{\pi}{6}+x-\frac{5 \pi}{6}-x\right)-\cos \left(\frac{\pi}{6}+x+\frac{5 \pi}{6}+x\right)\right)}{\cos ^{2} x}+\frac{2 \sin x}{\cos x}=0 \Leftrightarrow \\
& \Leftrightarrow \frac{2\left(\cos \frac{2}{3} \pi-\cos (\pi+2 x)\right)+2 \sin x \cos x}{\cos ^{2} x}=0 \Leftrightarrow \\
& \Leftrightarrow-1+2 \cos 2 x+2 \sin x \cos x=0 \Leftrightarrow \\
& \Leftrightarrow-\cos ^{2} x-\sin ^{2} x+2 \cos ^{2} x-2 \sin ^{2} x+2 \sin x \cos x=0 \Leftrightarrow \\
& \Leftrightarrow 3 \sin ^{2} x-2 \sin x \cos x-\cos ^{2} x=0 \Leftrightarrow 3 \operatorname{tg}^{2} x-2 \operatorname{tg} x-1=0
\end{aligned}
$$

Solving this equation as a quadratic in $\operatorname{tg} x$, we get $\operatorname{tg} x=-\frac{1}{3}, x_{1}=-\operatorname{arctg} \frac{1}{3}+\pi k, k \in Z, \operatorname{tg} x=1, x_{2}=\frac{\pi}{4}+\pi n=\frac{\pi}{4}(4 n+1), n \in Z$.

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