Let [| (60-) (60+)| (1)/(8), [0,2 ]].

Question

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

Accepted Answer

$$\begin{aligned}
& |\cos \theta \cos (60-\theta) \cos (60+\theta)| \leq \frac{1}{8} \
& \Rightarrow \frac{1}{4}|\cos 3 \theta| \leq \frac{1}{8} \
& \cos 3 \theta \text { is max if } \cos 3 \theta=\frac{1}{2} \
& \therefore \theta=\frac{\pi}{9}, \frac{5 \pi}{9}, \frac{7 \pi}{9}, \frac{11 \pi}{9}, \frac{13 \pi}{9}, \frac{17 \pi}{9} \
& \sum \theta_i=6 \pi
\end{aligned}$$

Solutions · 1