To solve the equation \( \sqrt{3} \tan(5x + \frac{\pi}{6}) = -3 \), we will first isolate the tangent function.

1. Divide both sides by \( \sqrt{3} \):
\[
\tan(5x + \frac{\pi}{6}) = -\frac{3}{\sqrt{3}} = -\sqrt{3}
\]

2. We know that \( \tan(\theta) = -\sqrt{3} \) corresponds to angles in the second and fourth quadrants where tangent is negative. The reference angle where \( \tan(\theta) = \sqrt{3} \) is \( \frac{\pi}{3} \).

Thus, the general solutions can be written as:
\[
5x + \frac{\pi}{6} = \frac{2\pi}{3} + n\pi \quad \text{(for the second quadrant)}
\]
\[
5x + \frac{\pi}{6} = -\frac{\pi}{3} + n\pi \quad \text{(for the fourth quadrant)}
\]
where \( n \) is any integer.

### Solving for \( x \)

**For the second quadrant:**
\[
5x + \frac{\pi}{6} = \frac{2\pi}{3} + n\pi
\]
Subtract \( \frac{\pi}{6} \) from both sides:
\[
5x = \frac{2\pi}{3} - \frac{\pi}{6} + n\pi
\]

To combine the terms, find a common denominator (here 6):
\[
\frac{2\pi}{3} = \frac{4\pi}{6}, \quad \text{thus,} \quad 5x = \frac{4\pi}{6} - \frac{\pi}{6} + n\pi = \frac{3\pi}{6} + n\pi = \frac{\pi}{2} + n\pi
\]

Now divide by 5:
\[
x = \frac{\pi}{10} + \frac{n\pi}{5}
\]

**For the fourth quadrant:**
\[
5x + \frac{\pi}{6} = -\frac{\pi}{3} + n\pi
\]
Subtract \( \frac{\pi}{6} \):
\[
5x = -\frac{\pi}{3} - \frac{\pi}{6} + n\pi
\]
Convert \( -\frac{\pi}{3} \):
\[
-\frac{\pi}{3} = -\frac{2\pi}{6}, \quad \text{thus,} \quad 5x = -\frac{2\pi}{6} - \frac{\pi}{6} + n\pi = -\frac{3\pi}{6} + n\pi = -\frac{\pi}{2} + n\pi
\]

Now divide by 5:
\[
x = -\frac{\pi}{10} + \frac{n\pi}{5}
\]

### Summary of Solutions
The complete solutions for \( x \) are:
\[
x = \frac{\pi}{10} + \frac{n\pi}{5} \quad \text{and} \quad x = -\frac{\pi}{10} + \frac{n\pi}{5}
\]
for any integer \( n \).