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 $n$.