To solve the equation $\sqrt{3}\tan(3\theta) + 1 = 0$ in the interval $[0, 2\pi)$, follow these steps:

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

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

Step 2: Identify the angles where $\tan(x) = -\frac{1}{\sqrt{3}}$

The tangent function is negative in the second and fourth quadrants. From the standard tangent values, $\tan(x) = -\frac{1}{\sqrt{3}}$ when: $x = \frac{5\pi}{6} \quad \text{and} \quad x = \frac{11\pi}{6}$

Thus: $3\theta = \frac{5\pi}{6} \quad \text{or} \quad 3\theta = \frac{11\pi}{6}$

Step 3: Solve for $\theta$

Divide by 3 to find $\theta$: $\theta = \frac{5\pi}{18} \quad \text{and} \quad \theta = \frac{11\pi}{18}$

Step 4: General solution for $\tan(x)$

The tangent function has a period of $\pi$, so additional solutions for $3\theta$ are: $3\theta = \frac{5\pi}{6} + n\pi \quad \text{and} \quad 3\theta = \frac{11\pi}{6} + n\pi \quad \text{where } n \in \mathbb{Z}$

Divide by 3: $\theta = \frac{5\pi}{18} + \frac{n\pi}{3} \quad \text{and} \quad \theta = \frac{11\pi}{18} + \frac{n\pi}{3}$

Step 5: Find solutions in $[0, 2\pi)$

Generate solutions for $n = 0, 1, 2, \dots$ until $\theta < 2\pi$:

1. For $\theta = \frac{5\pi}{18} + \frac{n\pi}{3}$:

   • $n = 0: \theta = \frac{5\pi}{18}$
   • $n = 1: \theta = \frac{5\pi}{18} + \frac{6\pi}{18} = \frac{11\pi}{18}$
   • $n = 2: \theta = \frac{5\pi}{18} + \frac{12\pi}{18} = \frac{17\pi}{18}$
   • $n = 3: \theta = \frac{5\pi}{18} + \frac{18\pi}{18} = \frac{23\pi}{18}$

2. For $\theta = \frac{11\pi}{18} + \frac{n\pi}{3}$:

   • $n = 0: \theta = \frac{11\pi}{18}$
   • $n = 1: \theta = \frac{11\pi}{18} + \frac{6\pi}{18} = \frac{17\pi}{18}$
   • $n = 2: \theta = \frac{11\pi}{18} + \frac{12\pi}{18} = \frac{23\pi}{18}$
   • $n = 3: \theta = \frac{11\pi}{18} + \frac{18\pi}{18} = \frac{29\pi}{18} > 2\pi$

Step 6: Final solutions

Combine the unique solutions within $[0, 2\pi)$: $\boxed{\frac{5\pi}{18}, \frac{11\pi}{18}, \frac{17\pi}{18}, \frac{23\pi}{18}}$

Let me know if you want further clarification or have any specific questions.

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Relative Questions:

1. How do you determine the quadrant where a trigonometric function is positive or negative?
2. Why does the tangent function have a period of $\pi$?
3. What are some common methods for solving trigonometric equations?
4. How do you convert between radians and degrees?
5. What other techniques can be used for equations involving square roots?

Tip: Always consider the periodicity of the trigonometric functions when solving for solutions in a specific interval.