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The given equation is:

\[
3 \cot^2(x) + 3 \cot(x) = 0
\]

We can solve this by factoring.

### Step 1: Factor out common terms
\[
3 \cot(x) \left( \cot(x) + 1 \right) = 0
\]

This gives two possible solutions:

1. \( \cot(x) = 0 \)
2. \( \cot(x) + 1 = 0 \)

### Step 2: Solve each equation

1. \( \cot(x) = 0 \)

   The cotangent function is zero when:
   \[
   x = \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z}
   \]

2. \( \cot(x) + 1 = 0 \)

   This simplifies to:
   \[
   \cot(x) = -1
   \]
   The cotangent function is -1 when:
   \[
   x = \frac{3\pi}{4} + n\pi, \quad n \in \mathbb{Z}
   \]

### Final Solution:
The general solutions for \(x\) are:
\[
x = \frac{\pi}{2} + n\pi \quad \text{or} \quad x = \frac{3\pi}{4} + n\pi, \quad n \in \mathbb{Z}
\]

Would you like more details on any step, or have questions about specific aspects?

Here are 5 related questions:
1. What is the domain of the cotangent function?
2. How would the solution change if the equation involved tangent instead of cotangent?
3. What are the periodic properties of the cotangent function?
4. How can we express \( \cot(x) \) in terms of sine and cosine?
5. Can this equation be solved graphically, and if so, how?

**Tip:** When solving trigonometric equations,always check the periodicity of the functions involved to ensure you capture all possible solutions across the domain of interest.

Step-by-step guide on solving the trigonometric equation 3cot(x)^2 + 3cot(x) = 0. Learn how to factor trigonometric equations and find the general solutions using the periodic properties of the cotangent function. Suitable for students in Grades 10-12.

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