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: Understanding the Concept:
To solve a trigonometric equation involving products of sine and cosine, we use double angle identities to simplify it to a standard form.
Step 2: Key Formula or Approach:
1. Double angle identity: \(\sin 2x = 2 \sin x \cos x\).
2. General solution for \(\sin \theta = \sin \alpha\) is \(\theta = n\pi + (-1)^n \alpha\), where \(n \in \mathbb{Z}\).
Step 3: Detailed Explanation:
Given:
\[ \sin x \cos x = \frac{1}{4} \]
Multiply both sides by 2:
\[ 2 \sin x \cos x = 2 \times \frac{1}{4} \]
\[ \sin 2x = \frac{1}{2} \]
We know that \(\sin \frac{\pi}{6} = \frac{1}{2}\).
So, \(\sin 2x = \sin \frac{\pi}{6}\).

Applying the general solution formula for sine:
\[ 2x = n\pi + (-1)^n \frac{\pi}{6} \]
Dividing the entire equation by 2:
\[ x = \frac{n\pi}{2} + (-1)^n \frac{\pi}{12} \]
Step 4: Final Answer:
The general solution is \(x = \frac{n\pi}{2} + (-1)^n \frac{\pi}{12}\).