The sum of the solutions in x in (0, 4pi) of | vedclass

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(C) Using the identity $\sin 	heta \sin(\frac{\pi}{3} - 	heta) \sin(\frac{\pi}{3} + 	heta) = \frac{1}{4} \sin(3	heta)$,we let $	heta = \frac{x}{3

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$3\pi$

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(C) Using the identity $\sin 	heta \sin(\frac{\pi}{3} - 	heta) \sin(\frac{\pi}{3} + 	heta) = \frac{1}{4} \sin(3	heta)$,we let $	heta = \frac{x}{3}$.Then the equation becomes $4 \sin \frac{x}{3} \sin(\frac{\pi}{3} + \frac{x}{3}) \sin(\frac{2\pi}{3} + \frac{x}{3}) = 1$.Note that $\sin(\frac{2\pi}{3} + \frac{x}{3}) = \sin(\pi - (\frac{\pi}{3} - \frac{x}{3})) = \sin(\frac{\pi}{3} - \frac{x}{3})$.Thus,the expression is $4 \sin \frac{x}{3} \sin(\frac{\pi}{3} + \frac{x}{3}) \sin(\frac{\pi}{3} - \frac{x}{3}) = 4 	imes \frac{1}{4} \sin(3 	imes \frac{x}{3}) = \sin x$.So,$\sin x = 1$.Given $x \in (0, 4\pi)$,the solutions are $x = \frac{\pi}{2}$ and $x = \frac{\pi}{2} + 2\pi = \frac{5\pi}{2}$.The sum of the solutions is $\frac{\pi}{2} + \frac{5\pi}{2} = \frac{6\pi}{2} = 3\pi$.

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

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