Find the general solution of the equation cos | vedclass

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Question

(A) Given equation: $\cos 4x = \cos 2x$$\Rightarrow \cos 4x - \cos 2x = 0$Using the formula $\cos A - \cos B = -2 \sin \left(\frac{A+B}{2}\right) \sin

Vedclass · Accepted Answer

$x = \frac{n\pi}{3}$ or $x = n\pi$,$n \in \mathbb{Z}$

Vedclass · Answer

(A) Given equation: $\cos 4x = \cos 2x$$\Rightarrow \cos 4x - \cos 2x = 0$Using the formula $\cos A - \cos B = -2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)$:$-2 \sin \left(\frac{4x+2x}{2}\right) \sin \left(\frac{4x-2x}{2}\right) = 0$$-2 \sin 3x \sin x = 0$$\sin 3x \sin x = 0$This implies $\sin 3x = 0$ or $\sin x = 0$.If $\sin 3x = 0$,then $3x = n\pi \Rightarrow x = \frac{n\pi}{3}$,where $n \in \mathbb{Z}$.If $\sin x = 0$,then $x = n\pi$,where $n \in \mathbb{Z}$.Combining these,the general solution is $x = \frac{n\pi}{3}$ or $x = n\pi$,where $n \in \mathbb{Z}$.

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

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