Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

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 $\left|\frac{1}{4}\cos 3\theta \right|\le \frac{1}{8}$ $\begin{array}{rl}& \Rightarrow |\cos 3\theta |\le \frac{1}{2}\\ & \Rightarrow -\frac{1}{2}\le \cos 3\theta \le \frac{1}{2}\end{array}$ $\Rightarrow$ maximum value of $\cos 3\theta =\frac{1}{2}$, here $\begin{array}{rl}& \Rightarrow 3\theta =2\mathrm{n}\pi \pm \frac{\pi }{3}\\ & \theta =\frac{2\mathrm{n}\pi }{3}\pm \frac{\pi }{9}\end{array}$ 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 $\frac{\pi }{9}+\frac{5\pi }{9}+\frac{7\pi }{9}+\frac{11\pi }{9}+\frac{13\pi }{9}+\frac{17\pi }{9}=\frac{54\pi }{9}=6\pi$