Verified Solution • 2025–26 Session moderate

QJEEM24Ma24bf: JEE Main 2024 Mathematics - Class 11 JEEM24Ma24bf Solution: Let

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

Multiple Choice Options

A 6π
B 9π
C 18π
D 15π

Step-by-Step Verified Solution

$\begin{aligned} | \cos θ \cos (60 - θ) \cos (60 + θ) | \leq \frac{1}{8} \\ \Rightarrow \frac{1}{4} | \cos 3 θ | \leq \frac{1}{8} \\ \cos 3 θ is max if \cos 3 θ = \frac{1}{2} \\ ∴ θ = \frac{π}{9}, \frac{5 π}{9}, \frac{7 π}{9}, \frac{11 π}{9}, \frac{13 π}{9}, \frac{17 π}{9} \\ \sum θ_{i} = 6 π \end{aligned}$

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