Let's solve the equation [tex]\(2 \sin 2\theta = 1\)[/tex] for [tex]\(0 \leq \theta \leq 360^\circ\)[/tex] step by step.
1. Start with the given equation:
[tex]\[
2 \sin 2\theta = 1.
\][/tex]
2. Isolate [tex]\(\sin 2\theta\)[/tex]:
[tex]\[
\sin 2\theta = \frac{1}{2}.
\][/tex]
3. Determine the general solutions for [tex]\(2\theta\)[/tex] when [tex]\(\sin \alpha = \frac{1}{2}\)[/tex]:
- The sine function equals [tex]\(\frac{1}{2}\)[/tex] at specific angles:
[tex]\[
2\theta = 30^\circ, \quad 150^\circ, \quad 390^\circ, \quad 510^\circ \quad (\text{since sine has a period of } 360^\circ).
\][/tex]
4. Solve for [tex]\(\theta\)[/tex] by dividing these angles by 2:
[tex]\[
\theta = \frac{30^\circ}{2} = 15^\circ,
\][/tex]
[tex]\[
\theta = \frac{150^\circ}{2} = 75^\circ,
\][/tex]
[tex]\[
\theta = \frac{390^\circ}{2} = 195^\circ,
\][/tex]
[tex]\[
\theta = \frac{510^\circ}{2} = 255^\circ.
\][/tex]
5. List all the solutions within the range [tex]\(0 \leq \theta \leq 360^\circ\)[/tex]:
[tex]\[
\theta = 15^\circ, \quad 75^\circ, \quad 195^\circ, \quad 255^\circ.
\][/tex]
Thus, the solutions for [tex]\(\theta\)[/tex] in the equation [tex]\(2 \sin 2\theta = 1\)[/tex] for [tex]\(0 \leq \theta \leq 360^\circ\)[/tex] are:
[tex]\[
\theta = 15^\circ, \quad 75^\circ, \quad 195^\circ, \quad 255^\circ.
\][/tex]
