To solve the equation \( \cos 3x = -\cos 7x \), we can rewrite it as:

\[
\cos 3x + \cos 7x = 0
\]

Using the identity for the sum of cosines, we can express \( \cos 3x + \cos 7x \) as:

\[
2 \cos\left(\frac{3x + 7x}{2}\right) \cos\left(\frac{3x - 7x}{2}\right) = 0
\]

This simplifies to:

\[
2 \cos(5x) \cos(-2x) = 0
\]

Since \( \cos(-2x) = \cos(2x) \), we can rewrite it as:

\[
2 \cos(5x) \cos(2x) = 0
\]

This means that either:

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

### Solving for \( \cos(5x) = 0 \)
The cosine function equals zero at odd multiples of \( \frac{\pi}{2} \):

\[
5x = \frac{\pi}{2} + n\pi \quad \text{for } n \in \mathbb{Z}
\]

So,

\[
x = \frac{\pi}{10} + \frac{n\pi}{5}
\]

### Solving for \( \cos(2x) = 0 \)
Similarly, we have:

\[
2x = \frac{\pi}{2} + m\pi \quad \text{for } m \in \mathbb{Z}
\]

This simplifies to:

\[
x = \frac{\pi}{4} + \frac{m\pi}{2}
\]

### Summary of Solutions
Thus the complete solution set for the equation \( \cos 3x = -\cos 7x \) is:

\[
x = \frac{\pi}{10} + \frac{n\pi}{5}, \quad n \in \mathbb{Z}
\]
\[
x = \frac{\pi}{4} + \frac{m\pi}{2}, \quad m \in \mathbb{Z}
\]

These represent all the angles \( x \) that satisfy the original equation.