The number of solutions of the equation cos l | vedclass

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Question

$\cos \left(\frac{\pi}{3}+x\right) \cos \left(\frac{\pi}{3}-x\right)=\frac{1}{4} \cos ^{2} 2 x$

$x \in[-3 \pi, 3 \pi]$

$4\left(\cos ^{2}\left(\f

Vedclass · Accepted Answer

$7$

Vedclass · Answer

$\cos \left(\frac{\pi}{3}+x\right) \cos \left(\frac{\pi}{3}-x\right)=\frac{1}{4} \cos ^{2} 2 x$

$x \in[-3 \pi, 3 \pi]$

$4\left(\cos ^{2}\left(\frac{\pi}{3}\right)-\sin ^{2} x\right)=\cos ^{2} 2 x$

$4\left(\frac{1}{4}-\sin ^{2} x\right)=\cos ^{2} 2 x$

$1-4 \sin ^{2} x=\cos ^{2} 2 x$

$1-2(1-\cos 2 x)=\cos ^{2} 2 x$

let $\cos 2 x = t$

$-1+2 \cos 2 x=\cos ^{2} 2 x$

$t ^{2}-2 t +1=0$

$( t -1)^{2}=0$

$t =1 \quad \cos 2 x =1$

$2 x =2 n \pi$

$x = n \pi$

$n =-3,-2,-1,0,1,2,3$

The number of solutions of the equation $\cos \left(x+\frac{\pi}{3}\right) \cos \left(\frac{\pi}{3}-x\right)=\frac{1}{4} \cos ^{2} 2 x, x \in[-3 \pi$ $3 \pi]$ is

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