The sum of solutions in x in (0, 2pi) of the | vedclass

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(A) Using the identity $\cos(60^{\circ} - x)\cos(60^{\circ} + x) = \cos^2(60^{\circ}) - \sin^2(x) = \frac{1}{4} - \sin^2(x)$.Substituting this into th

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(A) Using the identity $\cos(60^{\circ} - x)\cos(60^{\circ} + x) = \cos^2(60^{\circ}) - \sin^2(x) = \frac{1}{4} - \sin^2(x)$.Substituting this into the equation: $4\cos(x) \left(\frac{1}{4} - \sin^2(x)\right) = 1$.$4\cos(x) \left(\frac{1 - 4\sin^2(x)}{4}\right) = 1$.$\cos(x)(1 - 4\sin^2(x)) = 1$.$\cos(x)(1 - 4(1 - \cos^2(x))) = 1$.$\cos(x)(4\cos^2(x) - 3) = 1$.$4\cos^3(x) - 3\cos(x) = 1$.Using the identity $\cos(3x) = 4\cos^3(x) - 3\cos(x)$, we get $\cos(3x) = 1$.For $x \in (0, 2\pi)$, $3x \in (0, 6\pi)$.$3x = 2\pi, 4\pi$.$x = \frac{2\pi}{3}, \frac{4\pi}{3}$.The sum of the solutions is $\frac{2\pi}{3} + \frac{4\pi}{3} = \frac{6\pi}{3} = 2\pi$.

The sum of solutions in $x \in (0, 2\pi)$ of the equation $4\cos(x)\cos\left(\frac{\pi}{3} - x\right)\cos\left(\frac{\pi}{3} + x\right) = 1$ is equal to: (in $\pi$)

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