To solve the equation [tex]\(\sin\left(\frac{\theta}{2}\right) = -\frac{1}{2}\)[/tex] over all real values of [tex]\(\theta\)[/tex], we proceed as follows:
1. Understand the basic trigonometric equation:
We start with [tex]\(\sin x = -\frac{1}{2}\)[/tex]. We know that sine is negative in the third and fourth quadrants. The reference angle for [tex]\(\sin x = \frac{1}{2}\)[/tex] is [tex]\(\frac{\pi}{6}\)[/tex], thus the solutions for [tex]\(\sin x = -\frac{1}{2}\)[/tex] are:
[tex]\[
x = -\frac{\pi}{6} + 2k\pi \quad \text{and} \quad x = \frac{7\pi}{6} + 2k\pi
\][/tex]
for any integer [tex]\(k\)[/tex].
2. Apply the solutions to [tex]\(\sin\left(\frac{\theta}{2}\right)\)[/tex]:
Let [tex]\(x = \frac{\theta}{2}\)[/tex]. Substitute [tex]\(x\)[/tex] with [tex]\(\frac{\theta}{2}\)[/tex] in the general solutions:
[tex]\[
\frac{\theta}{2} = -\frac{\pi}{6} + 2k\pi \quad \text{and} \quad \frac{\theta}{2} = \frac{7\pi}{6} + 2k\pi
\][/tex]
3. Solve for [tex]\(\theta\)[/tex]:
Multiply both sides of each equation by 2 to solve for [tex]\(\theta\)[/tex]:
[tex]\[
\theta = -\frac{\pi}{3} + 4k\pi \quad \text{and} \quad \theta = \frac{7\pi}{3} + 4k\pi
\][/tex]
4. Rewrite the solutions:
Using [tex]\(n\)[/tex] as any integer (where [tex]\(n = k\)[/tex]):
[tex]\[
\theta = (-\frac{\pi}{3}) + 4n\pi \quad \text{and} \quad \theta = (\frac{7\pi}{3}) + 4n\pi
\][/tex]
5. Additional equivalent form:
Noting that [tex]\(\frac{7\pi}{3}\)[/tex] can be expressed as [tex]\(\frac{4\pi}{3} + \pi\)[/tex], we get another commonly used form:
[tex]\[
\theta = (\frac{4\pi}{3}) + 4n\pi
\][/tex]
Thus, the general solutions for [tex]\(\theta\)[/tex] are:
[tex]\[
\theta = (-\frac{\pi}{3}) + 4n\pi \quad \text{and} \quad \theta = (\frac{4\pi}{3}) + 4n\pi
\][/tex]
So the correct multiple-choice answer is:
[tex]\[
\theta = \frac{7\pi}{3}, \frac{11\pi}{3}
\][/tex]
