Question

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

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (A)

Given: $\sin 3x=\cos 2x$
$\Rightarrow 3\sin x-4{\sin }^{3}x=1-2{\sin }^{2}x$
$\Rightarrow 4{\sin }^{3}x-2{\sin }^{2}x-3\sin x+1=0$
$\Rightarrow 4{\sin }^{2}x(\sin x-1)+2\sin x(\sin x-1)-(\sin x-1)=0$
$\Rightarrow (\sin x-1)\left(4{\sin }^{2}x+2\sin x-1\right)$
$\Rightarrow \sin x=1$ or $4{\sin }^{2}x+2\sin x-1=0$
$4{\sin }^{2}x+2\sin x-1\sin x=\frac{-2\pm \sqrt{4+16}}{2.4}$
$4{\sin }^{2}x+2\sin x-1\sin x=\frac{-1\pm \sqrt{5}}{4}$
We know that, $\sin x>0\forall x\in (\pi /2,\pi )$ and $\sin x\ne 1$
$\sin x=\frac{\sqrt{5}-1}{4}$
Therefore, number of solution is 1 .