To find all solutions to the equation [tex]\((2 \sin x - 1)(2 \cos x + 1) = 0\)[/tex] in the interval [tex]\([0, 2\pi)\)[/tex], we need to solve both factors separately:
1. [tex]\(2 \sin x - 1 = 0\)[/tex]
2. [tex]\(2 \cos x + 1 = 0\)[/tex]
### Solving [tex]\(2 \sin x - 1 = 0\)[/tex]
1. Set [tex]\(2 \sin x - 1 = 0\)[/tex].
2. Solving for [tex]\(\sin x\)[/tex], we get:
[tex]\[
\sin x = \frac{1}{2}
\][/tex]
3. The general solutions for [tex]\(\sin x = \frac{1}{2}\)[/tex] are:
[tex]\[
x = \frac{\pi}{6} + 2k\pi \quad \text{and} \quad x = \pi - \frac{\pi}{6} + 2k\pi
\][/tex]
where [tex]\(k\)[/tex] is an integer.
4. In the interval [tex]\([0, 2\pi)\)[/tex], we find:
[tex]\[
x = \frac{\pi}{6} \quad \text{and} \quad x = \frac{5\pi}{6}
\][/tex]
### Solving [tex]\(2 \cos x + 1 = 0\)[/tex]
1. Set [tex]\(2 \cos x + 1 = 0\)[/tex].
2. Solving for [tex]\(\cos x\)[/tex], we get:
[tex]\[
\cos x = -\frac{1}{2}
\][/tex]
3. The general solutions for [tex]\(\cos x = -\frac{1}{2}\)[/tex] are:
[tex]\[
x = \frac{2\pi}{3} + 2k\pi \quad \text{and} \quad x = \frac{4\pi}{3} + 2k\pi
\][/tex]
where [tex]\(k\)[/tex] is an integer.
4. In the interval [tex]\([0, 2\pi)\)[/tex], we find:
[tex]\[
x = \frac{2\pi}{3} \quad \text{and} \quad x = \frac{4\pi}{3}
\][/tex]
### Combining Solutions
Thus, the solutions of the equation [tex]\((2 \sin x - 1)(2 \cos x + 1) = 0\)[/tex] in the interval [tex]\([0, 2\pi)\)[/tex] are:
[tex]\[
x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{2\pi}{3}, \frac{4\pi}{3}
\][/tex]
### Final Answer
[tex]\[
x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{2\pi}{3}, \frac{4\pi}{3}
\][/tex]
