To solve the equation \(\sin(2x) = -\frac{1}{2}\), we need to find the values of \(x\) that satisfy this equation. Here's how you can do it:

1. **Understand the Basic Equation:**
\[
\sin(2x) = -\frac{1}{2}
\]

2. **Use the General Solution for Sine:**
The general solution for \(\sin \theta = -\frac{1}{2}\) is:
\[
\theta = \arcsin(-\frac{1}{2}) + 2\pi n \quad \text{or} \quad \theta = \pi - \arcsin(-\frac{1}{2}) + 2\pi n
\]
for any integer \(n\).

3. **Calculate the Specific Solutions:**
The values of \(\theta\) that satisfy \(\sin \theta = -\frac{1}{2}\) are:
\[
\theta = -\frac{\pi}{6} + 2\pi n \quad \text{or} \quad \theta = \frac{7\pi}{6} + 2\pi n
\]
(since \(\arcsin(-\frac{1}{2}) = -\frac{\pi}{6}\)).

4. **Substitute Back for \(2x\):**
Since we have \(\sin(2x) = -\frac{1}{2}\), substitute \(\theta = 2x\).

This gives:
\[
2x = -\frac{\pi}{6} + 2\pi n \quad \text{or} \quad 2x = \frac{7\pi}{6} + 2\pi n
\]

5. **Solve for \(x\):**
Divide each equation by 2 to solve for \(x\):

\[
x = -\frac{\pi}{12} + \pi n \quad \text{or} \quad x = \frac{7\pi}{12} + \pi n
\]

6. **General Solutions:**
The general solutions for the original equation are:
\[
x = -\frac{\pi}{12} + \pi n \quad \text{and} \quad x = \frac{7\pi}{12} + \pi n
\]
where \(n\) is any integer.

These solutions give all \(x\) that satisfy \(\sin(2x) = -\frac{1}{2}\).