The number of solutions of the equation cos l | vedclass

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(D) Given equation: $\cos \left(\frac{\pi}{3}+x\right) \cos \left(\frac{\pi}{3}-x\right)=\frac{1}{4} \cos ^{2} 2 x$.Using the identity $\cos(A+B)\cos(

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$7$

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(D) Given equation: $\cos \left(\frac{\pi}{3}+x\right) \cos \left(\frac{\pi}{3}-x\right)=\frac{1}{4} \cos ^{2} 2 x$.Using the identity $\cos(A+B)\cos(A-B) = \cos^2 A - \sin^2 B$,we get:$\cos^2 \left(\frac{\pi}{3}\right) - \sin^2 x = \frac{1}{4} \cos^2 2x$.Substituting $\cos^2 \left(\frac{\pi}{3}\right) = \frac{1}{4}$:$\frac{1}{4} - \sin^2 x = \frac{1}{4} \cos^2 2x$.Multiplying by $4$:$1 - 4 \sin^2 x = \cos^2 2x$.Using $1 - 2 \sin^2 x = \cos 2x$,we have $4 \sin^2 x = 2(1 - \cos 2x)$:$1 - 2(1 - \cos 2x) = \cos^2 2x$.$1 - 2 + 2 \cos 2x = \cos^2 2x$.$\cos^2 2x - 2 \cos 2x + 1 = 0$.$(\cos 2x - 1)^2 = 0$.$\cos 2x = 1$.$2x = 2n\pi \implies x = n\pi$.For $x \in [-3\pi, 3\pi]$,the values of $n$ are $\{-3, -2, -1, 0, 1, 2, 3\}$.There are $7$ such values.

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

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