To solve the equation \( 3\cot(6x) + \sqrt{3} = 0 \), we can follow these steps:

1. **Isolate \(\cot(6x)\):**

\[
3\cot(6x) = -\sqrt{3}
\]

2. **Divide by 3:**

\[
\cot(6x) = -\frac{\sqrt{3}}{3}
\]

3. **Find the general solution for \( \cot(6x) = -\frac{\sqrt{3}}{3} \):**

   The cotangent of an angle \( \theta \) is \( -\frac{\sqrt{3}}{3} \) when \( \theta = \frac{5\pi}{6} \) or \( \theta = \frac{7\pi}{6} \), since cotangent is negative in the second and fourth quadrants.

   Therefore,

\[
6x = \frac{5\pi}{6} + n\pi \quad \text{or} \quad 6x = \frac{7\pi}{6} + n\pi \quad \text{for integer } n
\]

4. **Solve for \( x \):**

   Divide each equation by 6 to find \( x \):

\[
x = \frac{5\pi}{36} + \frac{n\pi}{6} \quad \text{or} \quad x = \frac{7\pi}{36} + \frac{n\pi}{6} \quad \text{for integer } n
\]

Thus, the general solutions are:

\[
x = \frac{5\pi}{36} + \frac{n\pi}{6} \quad \text{or} \quad x = \frac{7\pi}{36} + \frac{n\pi}{6} \quad \text{where } n \text{ is any integer}
\]

Would you like to go into more detail or have any questions about these steps?

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Here are five related questions to consider:

1. What are the general properties of the cotangent function?
2. How would the solution change if the equation were \( 3\cot(6x) - \sqrt{3} = 0 \)?
3. What are the key differences between solving equations involving cotangent and tangent functions?
4. How would you solve the equation if \( 6x \) had been replaced with \( 6x + \pi \)?
5. What happens if you apply a transformation to the angle, such as \( \cot(6x + \alpha) \) instead of \( \cot(6x) \)?

**Tip:** When dealing with trigonometric equations, always remember that solutions can often be periodic, so it's important to consider all possible angles within the given range.

Explore how to solve the trigonometric equation 3cot(6x) + sqrt(3) = 0 step-by-step. Learn about cotangent function properties and their applications in solving equations. Suitable for advanced high school students and beyond.

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