To solve the equation \( \sin x = \cos 5x \sin x \), we can start by rewriting the equation:

\[
\sin x - \cos 5x \sin x = 0
\]

Factoring out \( \sin x \) gives:

\[
\sin x (1 - \cos 5x) = 0
\]

This product equals zero if either factor is zero. Therefore, we consider two cases:

1. \( \sin x = 0 \)
2. \( 1 - \cos 5x = 0 \)

### Case 1: \( \sin x = 0 \)

The solutions for \( \sin x = 0 \) are:

\[
x = n\pi,
\]

where \( n \) is any integer.

### Case 2: \( 1 - \cos 5x = 0 \)

This simplifies to:

\[
\cos 5x = 1.
\]

The solutions for \( \cos 5x = 1 \) are:

\[
5x = 2k\pi,
\]

where \( k \) is any integer. Thus, we can divide by 5:

\[
x = \frac{2k\pi}{5}.
\]

### Final Solutions

Combining both cases, we have:

1. From \( \sin x = 0 \): \( x = n\pi \).
2. From \( \cos 5x = 1 \): \( x = \frac{2k\pi}{5} \).

So the complete solution set is given by:

\[
x = n\pi \quad \text{and} \quad x = \frac{2k\pi}{5}
\]

where \( n \) and \( k \) are integers.