
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
We have the following values

\[3[\sin 2x] \in \{-3, 0, 3\}, \quad 2[\cos x] \in \{-2, 0, 2\}, \quad [\sin 2x] \in \{-1, 0, 1\}.\]

To solve the equation, they provide only three options

\[
\begin{aligned}
&[\sin x] = -1, \ [\cos x] = 1, \ [\sin 2 x] = -1 \\
&[\sin x] = 0, \ [\cos x] = 0, \ [\sin 2 x] = 0 \\
&[\sin x] = 1, \ [\cos x] = -1, \ [\sin 2 x] = 1
\end{aligned}
\]

Consider the first option.

\[
\left\{\begin{array}{l}
{[\sin x] = -1} \\
{[\cos x] = 1} \\
{[\sin 2x] = -1}
\end{array}\right.\Leftrightarrow \left\{\begin{array}{l}
-1 \leq \sin x < 0 \\
\cos x = 1 \\
-1 \leq \sin 2x < 0
\end{array}\right.
\]

From the second equation, it follows that

\[
\sin x = 0
\]

And this contradicts the first equation. In this case, there are no solutions. Consider the second option.

\[
\left\{\begin{array}{l}
{[\sin x] = 0} \\
{[\cos x] = 0} \\
{[\sin 2 x] = 0}
\end{array}\right.\Leftrightarrow \left\{\begin{array}{l}
0 \leq \sin x < 1 \\
0 \leq \cos x < 1 \\
0 \leq \sin 2x < 1
\end{array}\right.
\]

From the first and second relations of the system, it follows that

\[
\sin 2x = 2 \sin x \cos x \geq 0
\]

And since \(\sin 2x \leq 1\), from the solutions of the first two double inequalities of the system, we only need to exclude points where \(\sin 2x = 1\). The solutions of the first two inequalities define the first quadrant on the trigonometric circle:

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