Let $|\cos \theta \cos (60-\theta) \cos (60+\theta)| \leq \frac{1}{8}, \theta \epsilon[0,2 \pi]$. Then, the sum of all $\theta \epsilon[0,2 \pi]$, where $\cos 3 \theta$ attains its maximum value, is :
$15 \pi$
$18 \pi$
$6 \pi$
$9 \pi$
Solution
We know that
$(\cos \theta)\left(\cos( 6 0 ^ { \circ } - \theta ) \left(\cos \left(60^{\circ}+\theta\right)=\frac{1}{4} \cos 3 \theta\right.\right.$
So equation reduces to $\left|\frac{1}{4} \cos 3 \theta\right| \leq \frac{1}{8}$
$\begin{aligned}
& \Rightarrow|\cos 3 \theta| \leq \frac{1}{2} \\
& \Rightarrow-\frac{1}{2} \leq \cos 3 \theta \leq \frac{1}{2}
\end{aligned}$
$\Rightarrow$ 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
$\frac{\pi}{9}+\frac{5 \pi}{9}+\frac{7 \pi}{9}+\frac{11 \pi}{9}+\frac{13 \pi}{9}+\frac{17 \pi}{9}=\frac{54 \pi}{9}=6 \pi$