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Ai批改: 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: $\boxed{2}$