Find the general solution of the equation cos | vedclass

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Question

$\cos 3 x+\cos x-\cos 2 x=0$

$ \Rightarrow 2\cos \left( {\frac{{3x + 2}}{2}} \right)\cos \left( {\frac{{3x - x}}{2}} \right) - \cos 2x = 0\quad $

Vedclass · Accepted Answer

Vedclass · Answer

$\cos 3 x+\cos x-\cos 2 x=0$

$ \Rightarrow 2\cos \left( {\frac{{3x + 2}}{2}} ight)\cos \left( {\frac{{3x - x}}{2}} ight) - \cos 2x = 0\quad $

$\left[ {\cos A + \cos B = 2\cos \left( {\frac{{A + B}}{2}} ight)\cos \left( {\frac{{A - B}}{2}} ight)} ight]$

$\Rightarrow 2 \cos 2 x \cos x-\cos 2 x=0$

$\Rightarrow \cos 2 x(2 \cos x-1)=0$

$\Rightarrow \cos 2 x=0 \quad$ or $\quad 2 \cos x-1=0$

$\Rightarrow \cos 2 x=0 \quad$ or $\quad \cos x=\frac{1}{2}$

$	herefore 2 x=(2 n+1) \frac{\pi}{2}$

or $\quad \cos x=\cos \frac{\pi}{3},$ where $n \in Z$

$\Rightarrow x=(2 n+1) \frac{\pi}{4}$

or $\quad x=2 n \pi \pm \frac{\pi}{3},$ where $n \in Z$

Find the general solution of the equation $\cos 3 x+\cos x-\cos 2 x=0$

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