Question

Find the general solution for each of the following equation:
$\cos 4x=\cos 2x$

• A. $x=\frac{n\pi }{2}orn\pi$
• B. $x=\frac{n\pi }{3}orn\pi$
• C. $x=2n\pi$
• D. $x=\frac{n\pi }{4}$

Answer 1

The correct option is B $x=\frac{n\pi }{3}orn\pi$

$\cos 4x=\cos 2x$

$\to \cos 4x-\cos 2x=0$.

we know that ,

$\cos A-\cos B=-2\sin \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)$

so, $\to -2\sin \left(\frac{4x+2x}{2}\right)\sin \left(\frac{4x-2x}{2}\right)=0$

$\to -2\sin 3x\sin x=0$

$\to \sin 3x\sin x=0$

either, $\sin 3x=0$

$3x=n\pi$

$x=\frac{n\pi }{3}$. ----------$1$

or, $\sin x=0$

$x=n\pi$ -----------$2$

from equation $1$ and $2$

option $B$ will be correct answer.