To solve the equation [tex]\( 2 \sin(x) - 1 = 0 \)[/tex] for [tex]\( x \)[/tex] within the interval [tex]\([0, 4\pi]\)[/tex], we can follow these detailed steps:
1. Start by isolating [tex]\(\sin(x)\)[/tex]:
[tex]\[
2 \sin(x) - 1 = 0
\][/tex]
Add 1 to both sides of the equation:
[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] within one period [tex]\( [0, 2\pi] \)[/tex]:
The sine function equals [tex]\(\frac{1}{2}\)[/tex] at the angles:
[tex]\[
x = \frac{\pi}{6} \quad \text{and} \quad x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}
\][/tex]
3. Extend the solutions to the interval [tex]\([0, 4\pi]\)[/tex]:
Since the sine function is periodic with period [tex]\(2\pi\)[/tex], we need to include all solutions within the interval [tex]\([0, 4\pi]\)[/tex]. We add multiples of the period [tex]\(2\pi\)[/tex] to the solutions found in the interval [tex]\([0, 2\pi]\)[/tex]:
[tex]\[
x = \frac{\pi}{6} + 2k\pi \quad \text{and} \quad x = \frac{5\pi}{6} + 2k\pi \quad \text{for integer } k
\][/tex]
Next, we find the specific solutions within the interval [tex]\([0, 4\pi]\)[/tex]:
For [tex]\( k=0 \)[/tex]:
[tex]\[
x_1 = \frac{\pi}{6}, \quad x_2 = \frac{5\pi}{6}
\][/tex]
For [tex]\( k=1 \)[/tex] (extending [tex]\(x\)[/tex] by one period [tex]\(2\pi\)[/tex]):
[tex]\[
x_3 = \frac{\pi}{6} + 2\pi = \frac{\pi}{6} + \frac{12\pi}{6} = \frac{13\pi}{6}
\][/tex]
[tex]\[
x_4 = \frac{5\pi}{6} + 2\pi = \frac{5\pi}{6} + \frac{12\pi}{6} = \frac{17\pi}{6}
\][/tex]
Check all solutions to ensure they lie within the interval [tex]\([0, 4\pi]\)[/tex]:
[tex]\[
0 \leq \frac{\pi}{6} < 4\pi, \quad 0 \leq \frac{5\pi}{6} < 4\pi, \quad 0 \leq \frac{13\pi}{6} < 4\pi, \quad 0 \leq \frac{17\pi}{6} < 4\pi
\][/tex]
The solutions within the given interval [tex]\([0, 4\pi]\)[/tex] are:
[tex]\[
x = \frac{\pi}{6}, \quad x = \frac{5\pi}{6}, \quad x = \frac{13\pi}{6}, \quad x = \frac{17\pi}{6}
\][/tex]
