To solve the equation \((1 + 2\cos x)(3 - \cos x) = 0\), we can set each factor to zero separately.

1. **First factor**:
\[
1 + 2\cos x = 0
\]
Solving for \(\cos x\):
\[
2\cos x = -1 \quad \Rightarrow \quad \cos x = -\frac{1}{2}
\]
The solutions for \(\cos x = -\frac{1}{2}\) are:
\[
x = \frac{2\pi}{3} + 2k\pi \quad \text{and} \quad x = \frac{4\pi}{3} + 2k\pi, \quad k \in \mathbb{Z}
\]

2. **Second factor**:
\[
3 - \cos x = 0
\]
Solving for \(\cos x\):
\[
\cos x = 3
\]
This is not possible since the range of \(\cos x\) is \([-1, 1]\).

**Final Solutions**:

The solutions to the original equation are:
\[
x = \frac{2\pi}{3} + 2k\pi \quad \text{and} \quad x = \frac{4\pi}{3} + 2k\pi, \quad k \in \mathbb{Z}
\]