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)

$5$

Answer 2

Given equation is $\sin x\cos 3x=\sin 3x\,\cos \,5x$
$\Rightarrow$ $2\sin x\,\cos \,3x-2\sin 3x\,cos5x=0$
$\Rightarrow$ $\sin (3x+x)-\sin (3x-x)-\sin (3x+5x)$
$+\sin (5x-3x)=0$
$\Rightarrow$ $\sin 4x-\sin 2x-\sin 8x+\sin 2x=0$
$\Rightarrow$ $\sin 4x-\sin 8x=0$
$\Rightarrow$ $2\cos \left( \frac{4x+8x}{2} \right)\sin \left( \frac{8x-4x}{2} \right)=0$
$\Rightarrow$ $2\cos \,6x\,\sin \,2x=0$
$\Rightarrow$ $\cos \,\,6x=0$ 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.