8.081. $6 \sin ^{2} x+\sin x \cos x-\cos ^{2} x=2$.
Solution.

## Given

$$
\begin{aligned}
& 6 \sin ^{2} x+\sin x \cos x-\cos ^{2} x-2\left(\sin ^{2} x+\cos ^{2} x\right)=0 \\
& 4 \sin ^{2} x+\sin x \cos x-3 \cos ^{2} x=0 \Leftrightarrow 4 \operatorname{tg}^{2} x+\operatorname{tg} x-3=0
\end{aligned}
$$

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

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