To solve the equation \( \sin(3x) + \cos\left(2x - \frac{\pi}{3}\right) = 0 \), we can rewrite the equation as:

\[
\sin(3x) = -\cos\left(2x - \frac{\pi}{3}\right)
\]

### Step 1: Expand the cosine term
Using the cosine angle addition formula, we have:

\[
\cos\left(2x - \frac{\pi}{3}\right) = \cos(2x)\cos\left(-\frac{\pi}{3}\right) - \sin(2x)\sin\left(-\frac{\pi}{3}\right)
\]
Knowing that \(\cos\left(-\frac{\pi}{3}\right) = \frac{1}{2}\) and \(\sin\left(-\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2}\):

\[
\cos\left(2x - \frac{\pi}{3}\right) = \frac{1}{2} \cos(2x) + \frac{\sqrt{3}}{2} \sin(2x)
\]

### Step 2: Substitute back into the equation
Now we substitute this back into the equation:

\[
\sin(3x) = -\left( \frac{1}{2} \cos(2x) + \frac{\sqrt{3}}{2} \sin(2x) \right)
\]

Multiplying through by \(-1\):

\[
-\sin(3x) = \frac{1}{2} \cos(2x) + \frac{\sqrt{3}}{2} \sin(2x)
\]

### Step 3: Analyze both sides
This equation can be approached graphically or numerically, but one analytic method to solve would be to express \(\sin(3x)\) in terms of \(\cos(2x)\) and \(\sin(2x)\).

### Step 4: Solve a simpler related equation
An alternative idea would be to seek solutions where both sides are equal to zero. One guess might be common angles.

For example, simple cases like:

- When \(3x = \frac{\pi}{2} + k\pi\) for integers \(k\)
- When \(2x - \frac{\pi}{3} = \frac{3\pi}{2} + m\pi\)

We can find \(x\) for these angles.

### Step 5: Solve for specific \(x\)

1. **From \(\sin(3x) = 0\)**

This gives:

\[
3x = n\pi \implies x = \frac{n\pi}{3}, \quad n \in \mathbb{Z}
\]

2. **From \(\cos(2x - \frac{\pi}{3}) = 0\)**

This gives:

\[
2x - \frac{\pi}{3} = \frac{\pi}{2} + m\pi \implies 2x = \frac{5\pi}{6} + m\pi \implies x = \frac{5\pi}{12} + \frac{m\pi}{2}, \quad m \in \mathbb{Z}
\]

### Conclusion
The final solutions can be obtained either numerically or graphically depending on the parity of the function involved. The solution sets would be obtained from both derived equations \(x = \frac{n\pi}{3}\) and \(x = \frac{5\pi}{12} + \frac{m\pi}{2}\).

Both families will yield solutions in the range of interest \(x \in \mathbb{R}\).

**Final results**:

\[
x = \frac{n\pi}{3}, \quad n \in \mathbb{Z}
\]
\[
x = \frac{5\pi}{12} + \frac{m\pi}{2}, \quad m \in \mathbb{Z}
\]

are the solutions to \( \sin(3x) + \cos(2x - \frac{\pi}{3}) = 0 \).