MCQ
Let $|\cos \theta \cos (60-\theta) \cos (60-\theta)| \leq \frac{1}{8}, \theta \in[0,2 \pi]$
Then, the sum of all $\theta \in[0,2 \pi]$, where $\cos 3 \theta$ attains its maximum value, is :
- A$9 \pi$
- B$18 \pi$
- ✓$6 \pi$
- D$15 \pi$
Then, the sum of all $\theta \in[0,2 \pi]$, where $\cos 3 \theta$ attains its maximum value, is :
$\cos \theta \cos ( 6 0^{ \circ } - \theta ) \cos (60^{\circ}+ \theta)=\frac{1}{4} \cos 3 \theta.$
So equation reduces to $\left|\frac{1}{4} \cos 3 \theta\right| \leq \frac{1}{8}$
$ \Rightarrow|\cos 3 \theta| \leq \frac{1}{2} $
$ \Rightarrow-\frac{1}{2} \leq \cos 3 \theta \leq \frac{1}{2}$
$\Rightarrow$ maximum value of $\cos 3 \theta=\frac{1}{2}$, here
$ \Rightarrow 3 \theta=2 \mathrm{n} \pi \pm \frac{\pi}{3} $
$ \theta=\frac{2 \mathrm{n} \pi}{3} \pm \frac{\pi}{9}$
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
$\frac{\pi}{9}+\frac{5 \pi}{9}+\frac{7 \pi}{9}+\frac{11 \pi}{9}+\frac{13 \pi}{9}+\frac{17 \pi}{9}=\frac{54 \pi}{9}=6 \pi$
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