JEE Main 2024 — Trigonometric Ratios & Identities Question with Solution

We know that $(\cos \theta )(\cos (60^{\circ }-\theta )(\cos \left(60^{\circ }+\theta \right)=\frac{1}{4}\cos 3\theta$ So equation reduces to $\left|\frac{1}{4}\cos 3\theta \right|\le \frac{1}{8}$ $\begin{aligned} & \Rightarrow|\cos 3 \theta| \leq \frac{1}{2} \\ & \Rightarrow-\frac{1}{2} \leq \cos 3 \theta \leq \frac{1}{2} \end{aligned}$ $\Rightarrow$ maximum value of $\cos 3\theta =\frac{1}{2}$, here $\begin{aligned} & \Rightarrow 3 \theta=2 \mathrm{n} \pi \pm \frac{\pi}{3} \\ & \theta=\frac{2 \mathrm{n} \pi}{3} \pm \frac{\pi}{9} \end{aligned}$ As $\theta \in [0,2\pi ]$ possible values are $\theta =\left\{\frac{\pi }{9},\frac{5\pi }{9},\frac{7\pi }{9},\frac{11\pi }{9},\frac{13\pi }{9},\frac{17\pi }{9}\right\}$ Whose sum is $\frac{\pi }{9}+\frac{5\pi }{9}+\frac{7\pi }{9}+\frac{11\pi }{9}+\frac{13\pi }{9}+\frac{17\pi }{9}=\frac{54\pi }{9}=6\pi$