Question

If sin $5x+\sin 3x+\sin x = 0$, then the value of x other than zero, lying between π$0\le \times \le \frac{\mathrm{\pi }}{2}$ is;

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

Answer 1

The Correct Option is C

Explanation:
Given:A trigonometric equation, $\sin 5x+\sin 3x+\sin x=0$,We have to find the value of $x$ other than zero lying between $0\le x\le \frac{\pi }{2}$Consider,$\sin 5x+\sin 3x+\sin x=0$$\Rightarrow (\sin x+\sin 5x)+\sin 3x=0$$\Rightarrow 2\sin 3x\cos 2x+\sin 3x=0[\because \sin C+\sin D=2\sin \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)]$$\Rightarrow \sin 3x(2\cos 2x+1)=0$$\Rightarrow \sin 3x=0$ or $2\cos 2x+1=0$But $\cos 2x\ne \frac{-1}{2}$ for $x\in [0,\frac{\pi }{2}]$Thus$\sin 3x=0$$\Rightarrow 3x=\pi$$\Rightarrow x=\frac{\pi }{3}$Hence, the correct option is (C).