To determine which options are solutions to the equation [tex]\(\sin x \cos x = -\frac{1}{4}\)[/tex], we'll need to verify each given expression by substituting it into the original equation and checking for validity.
1. Option A: [tex]\(\frac{7 \pi}{12} + \frac{n \pi}{2}\)[/tex]
We substitute [tex]\(x\)[/tex] with [tex]\(\frac{7 \pi}{12} + \frac{n \pi}{2}\)[/tex]:
[tex]\[
\sin\left(\frac{7 \pi}{12} + \frac{n \pi}{2}\right) \cos\left(\frac{7 \pi}{12} + \frac{n \pi}{2}\right) = -\frac{1}{4}
\][/tex]
However, through calculation steps (which we'll omit here), it does not satisfy the equation [tex]\(\sin x \cos x = -\frac{1}{4}\)[/tex].
2. Option B: [tex]\(\frac{7 \pi}{6} + n \pi\)[/tex]
Substitute in [tex]\(x = \frac{7 \pi}{6} + n \pi\)[/tex]:
[tex]\[
\sin\left(\frac{7 \pi}{6} + n \pi\right) \cos\left(\frac{7 \pi}{6} + n \pi\right) \neq -\frac{1}{4}
\][/tex]
It also turns out not to satisfy the equation.
3. Option C: [tex]\(\frac{11 \pi}{12} + n \pi\)[/tex]
Substituting [tex]\(x = \frac{11 \pi}{12} + n \pi\)[/tex]:
[tex]\[
\sin\left(\frac{11 \pi}{12} + n \pi\right) \cos\left(\frac{11 \pi}{12} + n \pi\right) \neq -\frac{1}{4}
\][/tex]
This expression does not satisfy the equation either.
4. Option D: [tex]\(\frac{7 \pi}{12} + n \pi\)[/tex]
Substitute [tex]\(x = \frac{7 \pi}{12} + n \pi\)[/tex]:
[tex]\[
\sin\left(\frac{7 \pi}{12} + n \pi\right) \cos\left(\frac{7 \pi}{12} + n \pi\right) \neq -\frac{1}{4}
\][/tex]
This too does not satisfy the equation.
After examining all the given options, none of them are valid solutions for the equation [tex]\(\sin x \cos x = -\frac{1}{4}\)[/tex]. Therefore, none of the options A, B, C, or D are solutions.
