470. $\sin ^{2} x+2 \sin x \cos x=3 \cos ^{2} x$.

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470. $\sin ^{2} x+2 \sin x \cos x=3 \cos ^{2} x$.
Solution. This is a homogeneous equation in terms of $\sin x$ and $\cos x$; dividing all its terms by $\cos ^{2} x \neq 0$, we get

$$
\frac{\sin ^{2} x}{\cos ^{2} x}+\frac{2 \sin x \cos x}{\cos ^{2} x}-\frac{3 \cos ^{2} x}{\cos ^{2} x}=0 ; \quad \operatorname{tg}^{2} x+2 \operatorname{tg} x-3=0
$$

Letting $\operatorname{tg} x=y$, we have the quadratic equation $y^{2}+2 y-3=0$. Solving it, we find $y_{1}=\frac{-2-4}{2}=-3, y_{2}=\frac{-2+4}{2}=1$.

Thus, $x_{1}=\operatorname{arctg}(-3)+\pi k=-\operatorname{arctg} 3+\pi k, x_{2}=\operatorname{arctg} 1+\pi k=\pi / 4+$ $+\pi k, k \in \mathbf{Z}$.