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

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(C) We use the identity $\cos 	heta \cos \left(\frac{\pi}{3}-	hetaight) \cos \left(\frac{\pi}{3}+	hetaight) = \frac{1}{4} \cos 3	heta$. Given

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$2\pi$

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(C) We use the identity $\cos 	heta \cos \left(\frac{\pi}{3}-	hetaight) \cos \left(\frac{\pi}{3}+	hetaight) = \frac{1}{4} \cos 3	heta$. Given the equation $\cos x \cos \left(\frac{\pi}{3}-xight) \cos \left(\frac{\pi}{3}+xight) = \frac{1}{4}$,we substitute the identity: $\frac{1}{4} \cos 3x = \frac{1}{4}$ $\cos 3x = 1$ For $x \in (0, 2\pi)$,we have $3x \in (0, 6\pi)$. The solutions for $\cos 3x = 1$ are $3x = 2\pi, 4\pi$. Thus,$x = \frac{2\pi}{3}, \frac{4\pi}{3}$. The sum of the solutions is $\frac{2\pi}{3} + \frac{4\pi}{3} = \frac{6\pi}{3} = 2\pi$.

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

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