3. (4 points) Solve the equation $\frac{\sin ^{2} 3 x}{\sin ^{2} x}=8 \cos 4 x+\frac{\cos ^{2} 3 x}{\cos ^{2} x}$.
Solution: O.D.Z. $x \neq \frac{\pi}{2} k, k \in Z$.

$$
\begin{aligned}
& \frac{\sin ^{2} 3 x}{\sin ^{2} x}=8 \cos 4 x+\frac{\cos ^{2} 3 x}{\cos ^{2} x} \Leftrightarrow \\
& \frac{\sin ^{2} 3 x \cos ^{2} x-\sin ^{2} x \cos ^{2} 3 x}{\sin ^{2} x \cos ^{2} x}=8 \cos 4 x \Leftrightarrow \frac{\sin 2 x \sin 4 x}{\frac{1}{4} \sin ^{2} 2 x}=8 \cos 4 x \Leftrightarrow \frac{8 \sin ^{2} 2 x \cos 2 x}{\sin ^{2} 2 x}=8 \cos 4 x
\end{aligned}
$$

$\Leftrightarrow$ (considering O.D.Z.) $\cos 2 x=\cos 4 x \Leftrightarrow-2 \sin 3 x \sin x=0$.

1) the roots of the equation $\sin x=0$ do not satisfy O.D.Z.;
2) $\sin 3 x=0 \Leftrightarrow x=\frac{\pi n}{3}, n \in Z$, but considering O.D.Z. $n \neq 3 k, k \in Z$, that is, the points marked with a cross on the graph are discarded.

![](https://cdn.mathpix.com/cropped/2024_05_06_0d1be90cfb2d2e1d4e02g-2.jpg?height=277&width=328&top_left_y=1438&top_left_x=1635)

Answer: $x= \pm \frac{\pi}{3}+\pi m, m \in Z$.