To solve the equation [tex]\(\sin(x)(\sin(x) - 1) = 0\)[/tex], we need to find the values of [tex]\(x\)[/tex] that satisfy this equation. Let's proceed step by step.
1. Understand the Equation:
[tex]\(\sin(x)(\sin(x) - 1) = 0\)[/tex] means that either [tex]\(\sin(x) = 0\)[/tex] or [tex]\(\sin(x) - 1 = 0\)[/tex].
2. Solve for [tex]\(\sin(x) = 0\)[/tex]:
The sine function [tex]\(\sin(x)\)[/tex] is zero at integer multiples of [tex]\(\pi\)[/tex]:
[tex]\[
x = k\pi \quad \text{where } k \in \mathbb{Z} \text{ (an integer)}
\][/tex]
3. Solve for [tex]\(\sin(x) - 1 = 0\)[/tex]:
Moving [tex]\(-1\)[/tex] to the other side, we get [tex]\(\sin(x) = 1\)[/tex]. The sine function [tex]\(\sin(x)\)[/tex] is 1 at:
[tex]\[
x = \frac{\pi}{2} + 2n\pi \quad \text{where } n \in \mathbb{Z} \text{ (an integer)}
\][/tex]
4. Combine the Solutions:
We have obtained two sets of solutions:
- From [tex]\(\sin(x) = 0\)[/tex]: [tex]\( x = k\pi \)[/tex]
- From [tex]\(\sin(x) = 1\)[/tex]: [tex]\( x = \frac{\pi}{2} + 2n\pi \)[/tex]
Combining these, we get the general solutions:
[tex]\[
x = k\pi \quad \text{and} \quad x = \frac{\pi}{2} + 2n\pi
\][/tex]
5. Verify with Given Choices:
Let's compare the derived solutions with the given options:
- A. [tex]\( x = \pm \pi n \)[/tex]
- B. [tex]\( x = \frac{\pi}{2} \pm 2\pi n, x = \frac{3\pi}{2} \pm 2\pi n \)[/tex]
- C. [tex]\( x = \pm \pi n, x = \frac{\pi}{2} \pm 2\pi n \)[/tex]
- D. [tex]\( x = \frac{\pi}{3} \pm 2\pi n, x = \frac{3\pi}{4} \pm 2\pi n \)[/tex]
The correct choice is the one that matches our derived set of solutions:
[tex]\[
x = k\pi \quad \text{and} \quad x = \frac{\pi}{2} + 2n\pi
\][/tex]
This corresponds to:
- [tex]\( x = \pm \pi n \)[/tex] (since [tex]\(k\)[/tex] is an integer, this general form matches [tex]\(k\pi\)[/tex])
- [tex]\( x = \frac{\pi}{2} \pm 2\pi n \)[/tex] (we can simplify the notation to match)
Therefore, the correct answer is:
C. [tex]\( x = \pm \pi n, x = \frac{\pi}{2} \pm 2\pi n \)[/tex]
