
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system

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

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

Hence, \(\cos x = 0\) or \(1 - 4 \sin ^{2} x = 0\).

It follows that \(x = \frac{\pi}{2} + \pi n\), \(x = \pm \frac{\pi}{6} + \pi k\).

Further, we have:

\[
\cos x \sin 3 x = 1
\]

This equality is possible in two cases:

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

It is easy to see that both systems have no solutions.

Present the answer in LaTex format: \boxed{Your answer}
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