To determine the number of solutions to the equation \( \tan(\theta) = \sqrt{3} \tan(\theta) \sqrt{3} \) in the interval \( [0, 2\pi) \), let's solve the equation step by step.

Step 1: Rearrange the equation
First, we rearrange the given equation to isolate terms involving \( \tan(\theta) \) on one side:
\[ \tan(\theta) = \sqrt{3} \tan(\theta) \sqrt{3} \]
Moving \( \sqrt{3} \tan(\theta) \) to the left side and adding \( \sqrt{3} \) to both sides gives:
\[ \tan(\theta) \sqrt{3} \tan(\theta) = \sqrt{3} \]
\[ \tan(\theta)(1 \sqrt{3}) = \sqrt{3} \]

2: Solve for \( \tan(\theta) \)
Now, solve for \( \tan(\theta) \):
\[ \tan(\theta) = \frac{\sqrt{3}}{1 \sqrt{3}} \]
To simplify, multiply the numerator and denominator by the conjugate of the denominator:
\[ \tan(\theta) = \frac{\sqrt{3}}{1 \sqrt{3}} \times \frac{1 + \sqrt{3}}{1 + \sqrt{3}} \]
\[ \tan(\theta) = \frac{\sqrt{3}(1 + \sqrt{3})}{1 3} \]
\[ \tan(\theta) = \frac{\sqrt{3} 3}{2} \]
\[ \tan(\theta) = \frac{\sqrt{3} + 3}{2} \]

3: Find \( \theta \) in the given interval
We need to find \( \theta \) in the interval \( [0, 2\pi) \) such that \( \tan(\theta) = \frac{\sqrt{3} + 3}{2} \).

4: Analyze the solution
Let \( k = \frac{\sqrt{3} + 3}{2} \)、This is a positive value since \( \sqrt{3} \approx 1.732 \), so \( k \approx \frac{1.732 + 3}{2} \approx 2.366 \).

5: Determine the number of solutions
The tangent function is periodic with period \( \pi \) and has a range of \( (\infty, \infty) \)、For any given value of \( \tan(\theta) = c \) where \( c \) is a real number, there are exactly two solutions in the interval \( [0, 2\pi) \) because the tangent function crosses any given value twice in its period, except at its asymptotes.

6: Consider the specific value
Given \( \tan(\theta) = \frac{\sqrt{3} + 3}{2} \), and knowing that \( \tan(\theta) \) is positive in the first and third quadrants, we can expect two solutions within \( [0, 2\pi) \) since \( \frac{\sqrt{3} + 3}{2} \) is a positive value.

7: Conclusion
The equation \( \tan(\theta) = \frac{\sqrt{3} + 3}{2} \) will have two solutions in the interval \( [0, 2\pi) \) because the tangent function is continuous and crosses any specific value (except at its vertical asymptotes) twice within any interval of length \( \pi \), and our interval \( [0, 2\pi) \) covers two full periods of \( \tan(\theta) \).

The final answer is: $\boxed{2}$