To solve the equation \( 2 \sin \theta - 1 = 0 \) in the interval \([0, 2\pi)\), follow these steps:
1. Isolate \(\sin \theta\):
[tex]\[
2 \sin \theta - 1 = 0
\][/tex]
Adding 1 to both sides, we get:
[tex]\[
2 \sin \theta = 1
\][/tex]
Dividing both sides by 2, we obtain:
[tex]\[
\sin \theta = \frac{1}{2}
\][/tex]
2. Find the principal solution:
Recall that the sine function equals \(\frac{1}{2}\) at certain standard angles. In the unit circle, this occurs at:
[tex]\[
\theta = \frac{\pi}{6}
\][/tex]
3. Consider the general solutions in the interval \([0, 2\pi)\):
The sine function is positive in the first and second quadrants. Therefore, the second angle where \(\sin \theta = \frac{1}{2}\) within \([0, 2\pi)\) is:
[tex]\[
\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}
\][/tex]
4. Write the solutions in terms of \(\pi\):
The solutions in the interval \([0, 2\pi)\) are:
[tex]\[
\theta = \frac{\pi}{6}, \frac{5\pi}{6}
\][/tex]
Therefore, the final solutions for \(\theta\) in radians in terms of \(\pi\) are:
[tex]\[
\theta = \frac{\pi}{6}, \frac{5\pi}{6}
\][/tex]
