To solve the equation [tex]\(2 \sin x = 1\)[/tex] over the interval [tex]\([0, 2\pi]\)[/tex], follow these steps:
1. Isolate the sine function:
[tex]\[
2 \sin x = 1
\][/tex]
Divide both sides by 2:
[tex]\[
\sin x = \frac{1}{2}
\][/tex]
2. Determine the general solutions for [tex]\(\sin x = \frac{1}{2}\)[/tex]:
Recall that the sine function equals [tex]\(\frac{1}{2}\)[/tex] for specific angles. From the unit circle, we know that:
[tex]\[
\sin x = \frac{1}{2} \text{ at } x = \frac{\pi}{6} + 2k\pi \text{ or } x = \pi - \frac{\pi}{6} + 2k\pi \text{ for integer } k
\][/tex]
Simplifying this:
[tex]\[
x = \frac{\pi}{6} + 2k\pi \quad \text{or} \quad x = \frac{5\pi}{6} + 2k\pi
\][/tex]
3. Find the specific solutions in the interval [tex]\([0, 2\pi]\)[/tex]:
We need to choose values of [tex]\(k\)[/tex] such that [tex]\( x \)[/tex] remains within the interval [tex]\([0, 2\pi]\)[/tex].
- For [tex]\( x = \frac{\pi}{6} + 2k\pi \)[/tex]:
- When [tex]\( k = 0 \)[/tex], [tex]\( x = \frac{\pi}{6} \)[/tex]
- When [tex]\( k = 1 \)[/tex], [tex]\( x = 2\pi + \frac{\pi}{6} = \frac{13\pi}{6} \)[/tex] (which is outside the interval [tex]\([0, 2\pi]\)[/tex])
- For [tex]\( x = \frac{5\pi}{6} + 2k\pi \)[/tex]:
- When [tex]\( k = 0 \)[/tex], [tex]\( x = \frac{5\pi}{6} \)[/tex]
- When [tex]\( k = 1 \)[/tex], [tex]\( x = 2\pi + \frac{5\pi}{6} = \frac{17\pi}{6} \)[/tex] (which is outside the interval [tex]\([0, 2\pi]\)[/tex])
4. List valid solutions:
Therefore, the values of [tex]\(x\)[/tex] that satisfy [tex]\(2 \sin x = 1\)[/tex] within the interval [tex]\([0, 2\pi]\)[/tex] are:
[tex]\[
x = \frac{\pi}{6}, \quad x = \frac{5\pi}{6}
\][/tex]
So the solutions to the equation [tex]\(2 \sin x = 1\)[/tex] over the interval [tex]\([0, 2\pi]\)[/tex] are:
[tex]\[
x = \frac{\pi}{6} \quad \text{and} \quad x = \frac{5\pi}{6}
\][/tex]
