JEE MAIN - Mathematics (2024 - 9th April Morning Shift - No. 6)
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 :
$$6 \pi$$
$$9 \pi$$
$$18 \pi$$
$$15 \pi$$
Explanation
$$\begin{aligned}
& |\cos \theta \cos (60-\theta) \cos (60+\theta)| \leq \frac{1}{8} \\
& \Rightarrow \frac{1}{4}|\cos 3 \theta| \leq \frac{1}{8} \\
& \cos 3 \theta \text { is max if } \cos 3 \theta=\frac{1}{2} \\
& \therefore \theta=\frac{\pi}{9}, \frac{5 \pi}{9}, \frac{7 \pi}{9}, \frac{11 \pi}{9}, \frac{13 \pi}{9}, \frac{17 \pi}{9} \\
& \sum \theta_i=6 \pi
\end{aligned}$$
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