Question

The number of solutions 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 $1$

$\sin 3x=\cos 2x$

$\Rightarrow$ $\sin 3x=\sin (\frac{\pi }{2}-2x)$

We know that,

$\sin A=\sin B$

$\Rightarrow$ $A=n\pi +(-1{)}^{n}B,$ where $n$ is an integer

Using the above identity, we get

$3x=n\pi +(-1{)}^n(\frac{\pi }{2}-2x)$

If $n=1,$ $x=\pi -\frac{\pi }{2}\Rightarrow x=\frac{\pi }{2}\notin (\frac{\pi }{2},\pi )$

If $n=2,$ $x=\frac{\pi }{2}\notin (\frac{\pi }{2},\pi )$

If $n=3,$ $x=\frac{5\pi }{2}\notin (\frac{\pi }{2},\pi )$

If $n=4,$ $x=\frac{9\pi }{10}\in (\frac{\pi }{2},\pi )$

If $n=5,$ $x=\frac{9\pi }{2}=\pi \notin (\frac{\pi }{2},\pi )$

Now, for all negative integers $x$ would e negative.

For all value of $n>5,$ solution $>\pi$

Hence, the only possible solution is for $n=4$ and $x=\frac{9\pi }{10}$

$\therefore$ The number of solutions of $\sin 3x=\cos 2x$ in the interval $(\frac{\pi }{2},\pi )$ is $1.$