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)

1

Answer 2

Given: $\sin 3 x=\cos 2 x$
$\Rightarrow 3 \sin x-4 \sin ^{3} x=1-2 \sin ^{2} x$
$\Rightarrow 4 \sin ^{3} x-2 \sin ^{2} x-3 \sin x+1=0$
$\Rightarrow 4 \sin ^{2} x(\sin x-1)+2 \sin x(\sin x-1)-(\sin x-1)=0$
$\Rightarrow (\sin x-1)\left(4 \sin ^{2} x+2 \sin x-1\right)$
$\Rightarrow \sin x=1$ 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 $\sin x \neq 1$
$\sin x=\frac{\sqrt{5}-1}{4}$
Therefore, number of solution is 1 .