We are tasked with solving the equation:

$\sqrt{3} \tan(3\theta) + 1 = 0$

in the interval $[0, 2\pi)$.

Steps to Solve:

1. Isolate $\tan(3\theta)$:

   $\sqrt{3} \tan(3\theta) = -1$

   $\tan(3\theta) = -\frac{1}{\sqrt{3}}$

2. Find the reference angle:

   Recall that $\tan(x) = -\frac{1}{\sqrt{3}}$ corresponds to an angle where the tangent is negative. Using the tangent inverse function for the positive ratio:

   $\tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$

3. Determine the quadrants:

   Since $\tan(3\theta)$ is negative, $3\theta$ lies in the second and fourth quadrants. The solutions for $3\theta$ are:

   $3\theta = \pi - \frac{\pi}{6}, \quad 3\theta = 2\pi - \frac{\pi}{6}$

   Simplify:

   $3\theta = \frac{5\pi}{6}, \quad 3\theta = \frac{11\pi}{6}$

4. Solve for $\theta$:

   Divide by 3 to isolate $\theta$:

   $\theta = \frac{5\pi}{18}, \quad \theta = \frac{11\pi}{18}$

5. Check for additional solutions in $[0, 2\pi)$:

   The period of $\tan(3\theta)$ is $\frac{\pi}{3}$. Adding multiples of this period will generate additional solutions. Since we are dividing the $3\theta$ interval into 3 parts:

   The general solutions are:

   $\theta = \frac{5\pi}{18}, \quad \theta = \frac{11\pi}{18}, \quad \theta = \frac{5\pi}{18} + \frac{\pi}{3}, \quad \theta = \frac{11\pi}{18} + \frac{\pi}{3}, \dots$

   These must be restricted to $[0, 2\pi)$. Calculating:

   $\frac{\pi}{3} = \frac{6\pi}{18}, \quad \frac{5\pi}{18} + \frac{6\pi}{18} = \frac{11\pi}{18}, \quad \dots$

   After simplifying, the valid solutions in $[0, 2\pi)$ are:

   $\theta = \frac{5\pi}{18}, \frac{11\pi}{18}, \frac{17\pi}{18}, \frac{23\pi}{18}$

Final Answer:

[ \boxed{\frac{5\pi}{18}, \frac{11\pi}{18}, \frac{17\pi}{18}, \frac{23\pi}{18}} \