# Task 3.

The domain of definition is $x \neq \frac{\pi K}{3}$. We have:

$$
\begin{gathered}
\cos ^{2} x=\sin ^{2} 2 x+\frac{\cos (2 x+x)}{\sin 3 x} \\
\cos ^{2} x=4 \sin ^{2} x \cos ^{2} x+\frac{\cos 2 x \cos x-\sin 2 x \sin x}{\sin 3 x} \\
\cos ^{2} x\left(1-4 \sin ^{2} x\right)=\frac{\cos x\left(\cos 2 x-2 \sin ^{2} x\right)}{\sin 3 x} \\
\cos ^{2} x\left(1-4 \sin ^{2} x\right)=\frac{\cos x}{\sin 3 x}\left(1-4 \sin ^{2} x\right)
\end{gathered}
$$

Thus, $\cos x=0$ or $1-4 \sin ^{2} x=0$.

From this, we find $x=\frac{\pi}{2}+\pi n, x= \pm \frac{\pi}{6}+\pi k$.

Further, we have:

$$
\cos x \sin 3 x=1
$$

This equality is possible in two cases:

$$
\left\{\begin{array} { c } 
{ \operatorname { cos } x = 1 } \\
{ \operatorname { sin } 3 x = 1 }
\end{array} \text { or } \left\{\begin{array}{c}
\cos x=-1 \\
\sin 3 x=-1
\end{array}\right.\right.
$$

It is easy to see that both systems of solutions have no solutions.
Answer: $\left\{\frac{\pi}{2}+\pi n ; \pm \frac{\pi}{6}+\pi k ; k, n \in \mathrm{Z}\right\}$.