Let's solve the given trigonometric equation step by step:
Given equation:
[tex]\[
\cot(x) \sin(x) + \cot(x) = 0
\][/tex]
First, recall the definition of the cotangent function:
[tex]\[
\cot(x) = \frac{\cos(x)}{\sin(x)}
\][/tex]
Substitute this identity into the original equation:
[tex]\[
\left(\frac{\cos(x)}{\sin(x)}\right) \sin(x) + \frac{\cos(x)}{\sin(x)} = 0
\][/tex]
Simplify the equation:
[tex]\[
\cos(x) + \frac{\cos(x)}{\sin(x)} = 0
\][/tex]
Factor out [tex]\(\cos(x)\)[/tex] from the left-hand side:
[tex]\[
\cos(x) \left( 1 + \frac{1}{\sin(x)} \right) = 0
\][/tex]
The product of two factors is zero if and only if at least one of the factors is zero. Therefore, we set each factor to zero:
Case 1:
[tex]\[
\cos(x) = 0
\][/tex]
The cosine function is zero at:
[tex]\[
x = \frac{\pi}{2} + n\pi, \quad \text{where } n \text{ is an integer}
\][/tex]
Case 2:
[tex]\[
1 + \frac{1}{\sin(x)} = 0
\][/tex]
Solving for [tex]\(\sin(x)\)[/tex]:
[tex]\[
\frac{1}{\sin(x)} = -1
\][/tex]
[tex]\[
\sin(x) = -1
\][/tex]
The sine function is -1 at:
[tex]\[
x = \frac{3\pi}{2} + 2n\pi, \quad \text{where } n \text{ is an integer}
\][/tex]
Now combine the solutions from both cases:
The solutions to the equation are:
[tex]\[
x = \frac{\pi}{2} + n\pi \quad \text{and} \quad x = \frac{3\pi}{2} + 2n\pi \quad \text{for integers } n
\][/tex]
