The equation provided is:

$3 \cot^2(x) + 3 \cot(x) = 0$

To solve this equation, we can factor it.

Step 1: Factor the equation

First, factor out the common factor, which is $3 \cot(x)$:

$3 \cot(x) [\cot(x) + 1] = 0$

Now, set each factor equal to zero:

1. $3 \cot(x) = 0$
2. $\cot(x) + 1 = 0$

Step 2: Solve each equation

1. $3 \cot(x) = 0$ $\cot(x) = 0$ The cotangent function is zero when $x = \frac{\pi}{2} + n\pi$, where $n$ is an integer.

2. $\cot(x) + 1 = 0$ $\cot(x) = -1$ The cotangent function equals $-1$ when $x = \frac{3\pi}{4} + n\pi$, where $n$ is an integer.

Final Solution

The general solutions for $x$ are:

• $x = \frac{\pi}{2} + n\pi$ for the first equation.
• $x = \frac{3\pi}{4} + n\pi$ for the second equation.

Thus, the solutions are:

$x = \frac{\pi}{2} + n\pi \quad \text{or} \quad x = \frac{3\pi}{4} + n\pi \quad \text{for integers} \, n.$

Would you like any more details or clarifications on this solution?

Here are some related questions to think about:

1. How is the cotangent function related to sine and cosine?
2. What is the period of the cotangent function, and how does it affect the general solution?
3. How can you derive the solutions for $\cot(x) = -1$ using reference angles?
4. What would happen if you had $3 \cot^2(x) + 3 = 0$ instead?
5. How would the solution change if the equation involved a different trigonometric function, like sine or cosine?

Tip: Always remember that trigonometric equations often have infinitely many solutions due to the periodic nature of trig functions.