Question

The number of solutions of $\sin 3x=\cos 2x,$ in the interval $(\frac{\pi }{2},\pi )$ is :

• A. $4$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is B $1$

$\sin 3x=\cos 2x\Rightarrow \cos 2x=\cos (\frac{\pi }{2}-3x)\Rightarrow 2x=2n\pi \pm (\frac{\pi }{2}-3x)$
$\Rightarrow 5x=(2n+\frac{1}{2})\pi ,-x=(2n-\frac{1}{2})\pi$
$\Rightarrow x=(4n+1)\frac{\pi }{10},x=-(4n-1)\frac{\pi }{2}$
But $x\in (\frac{\pi }{2},\pi )$
$\therefore x=\frac{9\pi }{10}$
Hence there is only one solution in $x\in (\frac{\pi }{2},\pi )$

Alternate solution:


The graph of $\sin 3x=\cos 2x$ is :


It is clear from the graph that the number of solutions in the interval $(\frac{\pi }{2},\pi )$ is $1$.