Question

The number of solutions of the equation $\sin x\cos 3x=\sin 3x\cos 5x$ in $\left[0,\frac{\pi }{2}\right]$ is

• A. $3$
• B. $4$
• C. $5$
• D. $6$

Answer 1

Answer: (C)

Given equation is $\sin x\cos 3x=\sin 3x\cos 5x$
$\Rightarrow$ $2\sin x\cos 3x-2\sin 3xcos5x=0$
$\Rightarrow$ $\sin (3x+x)-\sin (3x-x)-\sin (3x+5x)$
$+\sin (5x-3x)=0$
$\Rightarrow$ $\sin 4x-\sin 2x-\sin 8x+\sin 2x=0$
$\Rightarrow$ $\sin 4x-\sin 8x=0$
$\Rightarrow$ $2\cos \left(\frac{4x+8x}{2}\right)\sin \left(\frac{8x-4x}{2}\right)=0$
$\Rightarrow$ $2\cos 6x\sin 2x=0$
$\Rightarrow$ $\cos 6x=0$ or $\sin 2x=0$
$\Rightarrow$ $6x=(2n+1)\frac{\pi }{2}$ or $x=\frac{n\pi }{2}$
$\Rightarrow$ $x=(2n+1)\frac{\pi }{12}$ or $x=\frac{n\pi }{2}$
$\Rightarrow$ $x=0,\frac{\pi }{2},\frac{\pi }{12},\frac{3\pi }{12},\frac{5\pi }{12}\in \left[0,\frac{\pi }{2}\right]$
$\therefore$ Number of solutions is 5.