
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
Given that \(\sin 3x \neq 0\) and \(x \neq \frac{\pi k}{3}\):

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

Therefore, the conditions are:
\[
\left\{\begin{array}{c}
\cos x = 0 \\
1 - 4 \sin^2 x = 0
\end{array}\right. 
\]
or \(\sin 3x \cos x = -1\).

The solutions are: 
\[
x = \frac{\pi}{2} + \pi n, \quad x = \pm \frac{\pi}{6} + \pi k.
\]

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

These systems have no solutions.

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