Question

If \(\sin5x+\sin3x+\sin x=0\), 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:

When three sine terms are given, try combining the first and last using \(\sin A+\sin B\). This often creates a common factor.

Answer 1

The Correct Option is B

We are given \[ \sin5x+\sin3x+\sin x=0. \] First combine \[ \sin5x+\sin x. \] Using the identity: \[ \sin A+\sin B=2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right). \] Put \[ A=5x,\quad B=x. \] Then, \[ \sin5x+\sin x = 2\sin\left(\frac{5x+x}{2}\right) \cos\left(\frac{5x-x}{2}\right). \] \[ =2\sin3x\cos2x. \] Therefore, \[ \sin5x+\sin3x+\sin x=0 \] becomes \[ 2\sin3x\cos2x+\sin3x=0. \] Take \(\sin3x\) common: \[ \sin3x(2\cos2x+1)=0. \] So either \[ \sin3x=0 \] or \[ 2\cos2x+1=0. \] From \[ \sin3x=0, \] we get \[ 3x=n\pi. \] \[ x=\frac{n\pi}{3}. \] In the interval \[ 0\leq x\leq \frac{\pi}{2}, \] and \(x\neq 0\), the possible value is \[ x=\frac{\pi}{3}. \] Therefore, the required value is \[ \frac{\pi}{3}. \]