Question

If \(\sin 5x+\sin 3x+\sin x=0\), then the value of \(x\) other than zero lying between \(0\leq x\leq \frac{\pi}{2}\) is

• A. \(\frac{\pi}{6}\)
• B. \(\frac{\pi}{3}\)
• C. \(\frac{\pi}{12}\)
• D. \(\frac{\pi}{4}\)

Hint:

For sums of sine terms, first use \(\sin A+\sin B\) formula to simplify.

Answer 1

The Correct Option is B

Step 1: Given, \[ \sin 5x+\sin 3x+\sin x=0 \]

Step 2: Use identity: \[ \sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2} \] \[ \sin 5x+\sin x=2\sin 3x\cos 2x \]

Step 3: Substitute: \[ 2\sin 3x\cos 2x+\sin 3x=0 \] \[ \sin 3x(2\cos 2x+1)=0 \]

Step 4: Since \(x\neq 0\), take: \[ 2\cos 2x+1=0 \] \[ \cos 2x=-\frac{1}{2} \]

Step 5: In the given interval, this gives: \[ 2x=\frac{2\pi}{3} \] \[ x=\frac{\pi}{3} \] \[ \boxed{\frac{\pi}{3}} \]