8.083. $\sin ^{2} x-2 \sin x \cos x=3 \cos ^{2} x$.

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8.083. $\sin ^{2} x-2 \sin x \cos x=3 \cos ^{2} x$.
Solution.

Dividing both sides of the equation by $\cos ^{2} x \neq 0$, we have $\operatorname{tg}^{2} x-2 \operatorname{tg} x-3=0$.

Solving this equation as a quadratic equation in terms of $\operatorname{tg} x$, we find

$(\operatorname{tg} x)_{1}=-1, x_{1}=-\frac{\pi}{4}+\pi k, k \in Z ;(\operatorname{tg} x)_{2}=3 ; x_{2}=\operatorname{arctg} 3+\pi n, n \in Z$.

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