To solve the equation [tex]\(\sin(2x) = 0\)[/tex], follow this step-by-step process:
1. Identify the solutions to the inner function:
[tex]\[
\sin(2x) = 0
\][/tex]
The sine function equals zero at integer multiples of [tex]\(\pi\)[/tex]. Therefore, we can write:
[tex]\[
2x = n\pi \quad \text{for} \quad n \in \mathbb{Z}
\][/tex]
2. Solve for [tex]\(x\)[/tex]:
Divide both sides of the equation by 2 to isolate [tex]\(x\)[/tex]:
[tex]\[
x = \frac{n\pi}{2} \quad \text{for} \quad n \in \mathbb{Z}
\][/tex]
3. List the specific solutions:
To find particular values, consider:
- For [tex]\( n = 0 \)[/tex]:
[tex]\[
x = \frac{0 \cdot \pi}{2} = 0
\][/tex]
- For [tex]\( n = 1 \)[/tex]:
[tex]\[
x = \frac{1 \cdot \pi}{2} = \frac{\pi}{2}
\][/tex]
Thus, the specific solutions for [tex]\(x\)[/tex] within one period of the sine function (i.e., between 0 and [tex]\(\pi\)[/tex]) are:
[tex]\[
x = 0 \quad \text{and} \quad x = \frac{\pi}{2}
\][/tex]
Hence, [tex]\(x\)[/tex] can be [tex]\(0\)[/tex] or [tex]\(\frac{\pi}{2}\)[/tex] as specific solutions. The general solution, considering all integer multiples, can be written as:
[tex]\[
x = \frac{n\pi}{2} \quad \text{for} \quad n \in \mathbb{Z}.
\][/tex]
