Question

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

• A. \( \dfrac{n\pi}{2} + (-1)^n \dfrac{\pi}{12} \)
• B. \( \dfrac{n\pi}{2} + \dfrac{\pi}{4} \)
• C. \( n\pi \pm \dfrac{\pi}{6} \)
• D. \( \dfrac{n\pi}{2} + (-1)^n \dfrac{\pi}{6} \)

Hint:

Whenever products like \( \sin x \cos x \) appear, immediately think of: \[ 2\sin x \cos x = \sin 2x \] to simplify the equation.

Answer 1

The Correct Option is A

Concept: Use the double angle identity: \[ 2\sin x \cos x = \sin 2x \] This converts the given trigonometric equation into a standard form.

Step 1: Apply double angle identity. Given: \[ \sin x \cos x = \frac14 \] Multiplying both sides by \(2\): \[ 2\sin x \cos x = \frac12 \] Using: \[ 2\sin x \cos x = \sin 2x \] we get: \[ \sin 2x = \frac12 \]

Step 2: Find general solution of \( \sin 2x = \frac12 \). We know: \[ \sin \theta = \frac12 \] has solutions: \[ \theta = n\pi + (-1)^n\frac{\pi}{6} \] Putting: \[ \theta = 2x \] \[ 2x = n\pi + (-1)^n\frac{\pi}{6} \] Dividing by \(2\): \[ x = \frac{n\pi}{2} + (-1)^n\frac{\pi}{12} \]

Step 3: Final answer. \[ \boxed{ x = \frac{n\pi}{2} + (-1)^n\frac{\pi}{12} } \]