Find the general solution of the equation cos | vedclass

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Question

$\cos 4 x=\cos 2 x$

$\Rightarrow \cos 4 x-\cos 2 x=0$

$\Rightarrow-2 \sin \left(\frac{4 x+2 x}{2}\right) \sin \left(\frac{4 x-2 x}{2}\right)=0$

Vedclass · Accepted Answer

Vedclass · Answer

$\cos 4 x=\cos 2 x$

$\Rightarrow \cos 4 x-\cos 2 x=0$

$\Rightarrow-2 \sin \left(\frac{4 x+2 x}{2}ight) \sin \left(\frac{4 x-2 x}{2}ight)=0$

$\left[\because \cos A-\cos B=-2 \sin \left(\frac{A+B}{2}ight) \sin \left(\frac{A-B}{2}ight)ight]$

$\Rightarrow \sin 3 x \sin x=0$

$\Rightarrow \sin 3 x=0$ or $\sin x=0$

$	herefore 3 x=n \pi$

or $\quad \sin x=0$

$	herefore 3 x=n \pi$

or $x=n \pi,$ where $n \in Z$

$\Rightarrow x=\frac{n \pi}{3}$

or $x=n \pi,$ where $n \in Z$

Find the general solution of the equation $\cos 4 x=\cos 2 x$

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