
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
For all \( x \), the equality must hold:

\[
\begin{gathered}
f(x+5 \pi) = f(x), \text{ i.e.,} \\
\sin (n x + 5 \pi n) \cos \frac{6(x + 5 \pi)}{n + 1} = \sin n x \cos \frac{6 x}{n + 1}.
\end{gathered}
\]

For \( n = 0 \), this equality is satisfied, so we do not need to consider this value in further analysis.

Case 1: \( n \) is even. The equality simplifies to:

\[
\begin{aligned}
& \sin n x \cos \frac{6(x+5 \pi)}{n+1} - \sin n x \cos \frac{6 x}{n+1} \Leftrightarrow \\
& \sin n x \left(\cos \frac{6 x}{n+1} - \cos \frac{6 x + 30 \pi)}{n + 1}\right) = 0 \Leftrightarrow \\
& 2 \sin n x \cdot \sin \frac{6 x + 15 \pi}{n + 1} \cdot \sin \frac{15 \pi}{n + 1} = 0.
\end{aligned}
\]

Considering the zeros of the first two functions involved:

\[
\sin n x = 0 \Leftrightarrow n x = \pi k \quad (k \in \mathbb{Z}) \Rightarrow \frac{x}{\pi} = \frac{k}{n},
\]

\[
\sin \frac{6 x + 15 \pi}{n + 1} = 0 \Leftrightarrow \frac{6 x + 15 \pi}{n + 1} = \pi m \quad (m \in \mathbb{Z}) \Rightarrow \frac{x}{\pi} = \frac{m(n + 1) - 15}{6}.
\]

Their ratio \( \frac{x}{\pi} \) will be rational. Therefore, there exists a value \( x \) for which the functions do not zero out (e.g., \( x = \pi y \), where \( y \) is an irrational number). Hence, the condition is equivalent to:

\[
\sin \frac{15 \pi}{n + 1} = 0.
\]

From this, we find \( \frac{15 \pi}{n + 1} = \pi m \), \( m \in \mathbb{Z} \) or \( \frac{15}{n + 1} = m \), \( m \in \mathbb{Z} \). Therefore, \( n + 1 \) must be an odd divisor of 15:

\[
n + 1 \in \{1, -1, 3, -3, 5, -5, 15, -15\} \Rightarrow n \in \{0, -2, 2, -4, 4, -6, 14, -16\}.
\]

Case 2: \( n \) is odd. The equality simplifies to:

\[
\begin{aligned}
& -\sin n x \cos \frac{6(x + 5 \pi)}{n + 1} - \sin n x \cos \frac{6 x}{n + 1} \Leftrightarrow \\
& \sin n x \left(\cos \frac{6 x}{n + 1} + \cos \frac{6 x + 30 \pi}{n + 1}\right) = 0 \Leftrightarrow \\
& 2 \sin n x \cdot \cos \frac{6 x + 15 \pi}{n + 1} \cdot \cos \frac{15 \pi}{n + 1} = 0.
\end{aligned}
\]

Similarly, it can be seen that:

\[
\cos \frac{15 \pi}{n + 1} = 0.
\]

From this, we find:

\[
\frac{15 \pi}{n + 1} = \frac{\pi(2m + 1)}{2}, \quad m \in \mathbb{Z} \Rightarrow \frac{30}{n + 1} = 2m + 1, \quad m \in \mathbb{Z}.
\]

Thus, \( n + 1 \in \{2, -2, 6, -6, 10, -10, 30, -30\} \Rightarrow n \in \{1, -3, 5, -7, 9, -11, 29, -31\} \).

Combining the values found, we obtain the answer.

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