Question

If $\sin x \cos x = \frac{1}{4}$, then the general solution is:

• A. $x = \frac{n\pi}{2} + (-1)^n \frac{\pi}{12}$
• B. $x = n\pi + (-1)^n \frac{\pi}{12}$
• C. $x = \frac{n\pi}{2} + (-1)^n \frac{\pi}{6}$
• D. $x = n\pi + (-1)^n \frac{\pi}{6}$

Hint:

Always try to convert products of sine and cosine into a single trigonometric function using double angle identities.

Answer 1

The Correct Option is A

Step 1: Concept
Use the double angle formula for sine: $\sin(2x) = 2 \sin x \cos x$.

Step 2: Meaning
Multiplying both sides of the given equation by 2:
$2 \sin x \cos x = 2 \times \frac{1}{4}$
$\sin(2x) = \frac{1}{2}$

Step 3: Analysis
The principal value for $\sin \theta = \frac{1}{2}$ is $\theta = \frac{\pi}{6}$. The general solution for $\sin \theta = \sin \alpha$ is $\theta = n\pi + (-1)^n \alpha$.
Substituting $\theta = 2x$ and $\alpha = \frac{\pi}{6}$:
$2x = n\pi + (-1)^n \frac{\pi}{6}$

Step 4: Conclusion
Dividing by 2 to isolate $x$:
$x = \frac{n\pi}{2} + (-1)^n \frac{\pi}{12}$ Final Answer: (A)