To determine which of the given options solve the equation \(\sin x \cos x = \frac{\sqrt{3}}{4}\), we can start by using a trigonometric identity and checking each option step-by-step.
First, let's use the double-angle identity for sine, which states:
[tex]\[
\sin(2x) = 2 \sin(x) \cos(x)
\][/tex]
We can rewrite the given equation using this identity:
[tex]\[
\sin x \cos x = \frac{\sqrt{3}}{4} \Rightarrow 2 \sin x \cos x = \frac{\sqrt{3}}{2} \Rightarrow \sin(2x) = \frac{\sqrt{3}}{2}
\][/tex]
Next, we need to determine the angles for which \(\sin(2x) = \frac{\sqrt{3}}{2}\).
The value \(\frac{\sqrt{3}}{2}\) corresponds to angles:
[tex]\[
2x = \frac{\pi}{3} + 2k\pi \quad \text{or} \quad 2x = \pi - \frac{\pi}{3} + 2k\pi
\][/tex]
where \(k\) is any integer.
Simplifying these equations gives:
[tex]\[
2x = \frac{\pi}{3} + 2k\pi \quad \text{or} \quad 2x = \frac{2\pi}{3} + 2k\pi
\][/tex]
Solving for \(x\):
[tex]\[
x = \frac{\pi}{6} + k\pi \quad \text{or} \quad x = \frac{\pi}{3} + k\pi
\][/tex]
where \(k\) is any integer.
Now we check each given option to see if they match either \(x = \frac{\pi}{6} + k\pi\) or \(x = \frac{\pi}{3} + k\pi\):
A. \(\frac{\pi}{12} + n\pi\)
This does not match. Simplifying it further does not yield either \(\frac{\pi}{6} + k\pi\) or \(\frac{\pi}{3} + k\pi\).
B. \(\frac{\pi}{12} + \frac{n\pi}{2}\)
This also does not match. Simplifying it further does not yield either \(\frac{\pi}{6} + k\pi\) or \(\frac{\pi}{3} + k\pi\).
C. \(\frac{2\pi}{6} + n\pi\)
Simplifying this gives:
[tex]\[
\frac{2\pi}{6} = \frac{\pi}{3}
\][/tex]
Thus, \(x = \frac{\pi}{3} + n\pi\), which matches one of our derived solutions.
D. \(\frac{\pi}{6} + n\pi\)
This matches exactly one of our derived solutions.
Therefore, the correct options are:
[tex]\[
\boxed{C \text{ and } D}
\][/tex]
Choices C ([tex]\(\frac{2\pi}{6} + n\pi\)[/tex]) and D ([tex]\(\frac{\pi}{6} + n\pi\)[/tex]) are indeed correct solutions to the equation [tex]\(\sin x \cos x = \frac{\sqrt{3}}{4}\)[/tex].
