Let |cos theta cos (60^{circ}-theta) cos (60^ | vedclass

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(C) We know the identity: $\cos 	heta \cos (60^{\circ} - 	heta) \cos (60^{\circ} + 	heta) = \frac{1}{4} \cos 3	heta$.The given inequality reduces

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$6$

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(C) We know the identity: $\cos 	heta \cos (60^{\circ} - 	heta) \cos (60^{\circ} + 	heta) = \frac{1}{4} \cos 3	heta$.The given inequality reduces to: $|\frac{1}{4} \cos 3	heta| \leq \frac{1}{8}$.This implies: $|\cos 3	heta| \leq \frac{1}{2}$, or $-\frac{1}{2} \leq \cos 3	heta \leq \frac{1}{2}$.The maximum value of $\cos 3	heta$ in this range is $\frac{1}{2}$.Solving $\cos 3	heta = \frac{1}{2}$ for $3	heta \in [0, 6\pi]$:$3	heta = 2n\pi \pm \frac{\pi}{3}$.For $n=0$: $3	heta = \frac{\pi}{3} \Rightarrow 	heta = \frac{\pi}{9}$.For $n=1$: $3	heta = 2\pi \pm \frac{\pi}{3} \Rightarrow 	heta = \frac{5\pi}{9}, \frac{7\pi}{9}$.For $n=2$: $3	heta = 4\pi \pm \frac{\pi}{3} \Rightarrow 	heta = \frac{11\pi}{9}, \frac{13\pi}{9}$.For $n=3$: $3	heta = 6\pi - \frac{\pi}{3} \Rightarrow 	heta = \frac{17\pi}{9}$.The sum of these values is: $\frac{\pi}{9} + \frac{5\pi}{9} + \frac{7\pi}{9} + \frac{11\pi}{9} + \frac{13\pi}{9} + \frac{17\pi}{9} = \frac{54\pi}{9} = 6\pi$.

Let $|\cos \theta \cos (60^{\circ}-\theta) \cos (60^{\circ}+\theta)| \leq \frac{1}{8}$, where $\theta \in [0, 2\pi]$. Then, the sum of all $\theta \in [0, 2\pi]$ where $\cos 3\theta$ attains its maximum value is: (in $\pi$)

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