To solve the equation [tex]\(\sin 2x + \frac{\sqrt{3}}{2} = 0\)[/tex]:
1. Isolate the trigonometric function:
[tex]\[
\sin 2x = -\frac{\sqrt{3}}{2}
\][/tex]
2. Determine the principal solutions:
The sine function takes the value [tex]\(-\frac{\sqrt{3}}{2}\)[/tex] at specific angles. We need to find [tex]\(2x\)[/tex] such that:
[tex]\[
\sin 2x = -\frac{\sqrt{3}}{2}
\][/tex]
This occurs at:
[tex]\[
2x = \frac{4\pi}{3} + 2k\pi \quad \text{and} \quad 2x = \frac{5\pi}{3} + 2k\pi
\][/tex]
where [tex]\(k\)[/tex] is any integer.
3. Solve for [tex]\(x\)[/tex]:
Divide each equation by 2:
[tex]\[
x = \frac{2\pi}{3} + k\pi \quad \text{and} \quad x = \frac{5\pi}{6} + k\pi
\][/tex]
4. Express the general solutions:
The general solutions for [tex]\(x\)[/tex] are:
[tex]\[
x = \frac{2\pi}{3} + k\pi \quad \text{and} \quad x = \frac{5\pi}{6} + k\pi
\][/tex]
To match the required format [tex]\(\frac{a\pi}{b} + c\pi n\)[/tex]:
[tex]\[
x = \frac{2\pi}{3} + \pi n \quad \text{and} \quad x = \frac{5\pi}{6} + \pi n
\][/tex]
5. Select the appropriate format from the given options:
Comparing the solutions:
- The first term matches [tex]\(\frac{2\pi}{3}\)[/tex] plus a multiple of [tex]\(\pi\)[/tex].
- The second term matches [tex]\(\frac{5\pi}{6}\)[/tex] plus a multiple of [tex]\(\pi\)[/tex].
The correct answer is:
[tex]\[
\frac{2\pi}{3} + \pi n \quad ; \quad \frac{5\pi}{6} + \pi n
\][/tex]
Thus, the correct option is:
- [tex]\(\frac{2\pi}{3} + \pi n \ ; \ \frac{5\pi}{6} + \pi n\)[/tex]
