Question

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

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

Simplifying $\sin 3x=\cos 2x$.

$\sin 3x=\cos 2x$

$3\sin x-4{\sin }^{3}x=1-2{\sin }^{2}x$

$4{\sin }^{3}x-2{\sin }^{2}x-3\sin x+1=0$

Put $\sin x=t$.

$4t^3-2t^2-3t+1=0$

$\left(t-1\right)\left(4t^2+2t-1\right)=0$

$t=1$

Or,

$4t^2+2t-1=0$

$t=\frac{-2\pm \sqrt{{\left(2\right)}^2-4\left(4\right)\left(-1\right)}}{2\times 4}$

$=\frac{-2\pm \sqrt{20}}{8}$

$=\frac{-2\pm 2\sqrt{5}}{8}$

$=\frac{-1\pm \sqrt{5}}{4}$

Then,

$\sin x=1$

$x=\frac{\pi }{2}$

Or,

$\sin x=\frac{-1\pm \sqrt{5}}{4}$

$\sin x=\frac{-1+\sqrt{5}}{4}$

$x=\pi +\frac{\pi }{10}$

$x=\frac{11\pi }{10}$

Or,

$\sin x=\frac{-1-\sqrt{5}}{4}$

$x=2\pi -\frac{\pi }{10}$

$x=\frac{19\pi }{10}$

So, $x=\frac{\pi }{2}$, $x=\frac{11\pi }{10}$ and $x=\frac{19\pi }{10}$.

Since the given interval is $\left(\frac{\pi }{2},\pi \right)$, then, $x=\frac{\pi }{2}$ is the only solution.

Therefore, the number of solutions of $\sin 3x=\cos 2x$ is 1.