Question

The number of solutions to the equation $sin5x+sin3x+sinx=0$ for $0\le x\le \pi$ is

• A. 1
• B. 2
• C. 3
• D. None of these

Answer 1

The correct option is D None of these

$sinx+sin3x+sin5x=0x\in [0,\pi ]$

$(sinx+sin5x)+sin3x=0$

$2sin3xcos2x+sin3x=0$ $[\because sinx+siny=2\left|sin\frac{x+y}{2}\right|\left|cos\frac{x-y}{2}\right|]$

$sin3x(2cos2x+1)=0$

$\Rightarrow sin3x=0,2cos2x+1=0$

$3x=n\pi$ $|$ , $cos2x=\frac{-1}{2}$

$x=\frac{n\pi }{3}$$|$ $2x=2n\pi \pm \frac{2\pi }{3}[\because cos2\frac{2\pi }{3}=\frac{-1}{2}]$

$|x=n\pi \pm \frac{\pi }{3}$

For $x\in [0,\pi ],x=0,\frac{\pi }{3},\frac{2\pi }{3},\pi$ [ 4 solutions]

$\therefore$ None of these is correct option