Question

The number of solutions of the equation $\left|\sin x\right|=\left|\cos 3x\right|$ in $[-2\pi ,2\pi ]$ is:

• A. $32$
• B. $28$
• C. $24$
• D. $30$

Answer 1

Answer: (C)

$24$

Given, $|\sin x|=|\cos 3x|$
$\sin x=\pm \cos 3x$

$\sin \left(x\right)=\cos 3x$ and $\sin x=-\cos 3x$

Now

$\sin x=\cos 3x$ gives us

$\sin x-\cos 3x=0$

$\sin x-\sin (\frac{\pi }{2}-3x)=0$

$2\cos (\frac{\pi }{4}-x)\sin (2x-\frac{\pi }{4})=0$

$2x-\frac{\pi }{4}=n\pi$

$x=\frac{n\pi }{2}-\frac{\pi }{8}$

$nϵ[-2,2]$

And

$\cos (x-\frac{\pi }{4})=0$

$x-\frac{\pi }{4}=n\pi +\frac{\pi }{2}$

$x=n\pi +\frac{\pi }{2}+\frac{\pi }{4}$

$=n\pi +\frac{3\pi }{4}$ ...$nϵ[-2,2]$.

Consider $\sin x=-\cos 3x$

$\sin x+\cos 3x=0$

$\sin x+\sin (\frac{\pi }{2}-3x)=0$

$2\sin (\frac{\pi }{4}-x)\cos (2x-\frac{\pi }{4})=0$

Hence, $\frac{\pi }{4}-x=n\pi$

$x=\frac{\pi }{4}-n\pi$

And

$2x-\frac{\pi }{4}=n\pi +\frac{\pi }{2}$

$2x=n\pi +\frac{3\pi }{4}$

$x=\frac{n\pi }{2}+\frac{3\pi }{8}$.....$nϵ[-2,2]$

Hence in total we get 24 solutions.