We are tasked with solving the equation $\cot(x) \sin(x) + \cot(x) = 0$.

Step 1: Factor the expression

We can factor out $\cot(x)$ from the equation:

$\cot(x) (\sin(x) + 1) = 0$

Step 2: Solve for each factor

Now, we solve for each factor separately.

Case 1: $\cot(x) = 0$

The cotangent function, $\cot(x)$, is zero when:

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

Case 2: $\sin(x) + 1 = 0$

This implies that $\sin(x) = -1$, which happens at:

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

Step 3: Combine solutions

The solutions to the equation come from both cases:

1. $x = \frac{\pi}{2} + n\pi$
2. $x = \frac{3\pi}{2} + 2n\pi$

Step 4: Identify the correct option

Comparing these solutions to the choices provided in the image, the correct answer matches:

$x = \frac{\pi}{2} + n\pi \quad \text{and} \quad x = \frac{3\pi}{2} + 2n\pi$

This corresponds to the second option:
$\boxed{\frac{\pi}{2} + n\pi, \quad \frac{3\pi}{2} + 2n\pi}$

Would you like more details, or do you have any questions? Here are five related questions to expand on this:

1. What is the general form for solving trigonometric equations involving cotangent?
2. How can we interpret the periodicity of trigonometric functions in solutions?
3. Why does $\cot(x)$ have zeros at $\frac{\pi}{2} + n\pi$?
4. What happens when you apply similar techniques to secant or cosecant equations?
5. How does factoring help in solving complex trigonometric equations?

Tip: Always check the domain of trigonometric functions, as they can introduce restrictions in certain cases.