# Task 3.

Domain of definition: $\sin 3 x \neq 0, x \neq \frac{\pi k}{3}$.

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

Therefore, $\left[\begin{array}{c}\cos x=0 \\ 1-4 \sin ^{2} x=0\end{array}\right.$ or $\sin 3 x \cos x=-1$.

Thus, the solution will be: $x=\frac{\pi}{2}+\pi n, x= \pm \frac{\pi}{6}+\pi k$.

Further, we have:

\[
\left\{\begin{array} { c } 
{ \operatorname { s i n } 3 x = - 1 } \\
{ \operatorname { c o s } x = 1 }
\end{array} \text { or } \left\{\begin{array}{l}
\sin 3 x=1 \\
\cos x=-1
\end{array}\right.\right.
\]

These systems have no solutions.
Answer: $\left\{\frac{\pi}{2}+\pi n, \pm \frac{\pi}{6}+\pi k, k, n \in Z\right\}$.