the sum of all where attains its maximum value, is Trigonometric equation

Question

Let $| \cos θ \cos (60 - θ) \cos (60 + θ) |$ $\leq \frac{1}{8},$ $θ \in [0, 2 π] .$ Then, the sum of all $θ \in [0, 2 π],$ where $\cos 3 θ$ attains its maximum value, is: [2024]

Smart Achievers · Accepted Answer

(3)

$| \cos θ \cos (60 - θ) \cos (60 + θ) |$ $\leq \frac{1}{8}$

$\Rightarrow$ $| \cos θ \cos (\frac{π}{3} - θ) \cos (\frac{π}{3} + θ) |$ $\leq \frac{1}{8}$

$\Rightarrow \frac{1}{4} | \cos 3 θ | \leq \frac{1}{8} \Rightarrow | \cos 3 θ | \leq \frac{1}{2} \Rightarrow \frac{- 1}{2} \leq \cos 3 θ \leq \frac{1}{2}$

$cos 3 θ is max if \cos 3 θ = \frac{1}{2}$

$\Rightarrow 3 θ = 2 n π \pm \frac{π}{3} \Rightarrow θ = \frac{2 n π}{3} \pm \frac{π}{9}$

when $θ \in [0, 2 π]$

$∴$ $θ = \frac{π}{9}, \frac{5 π}{9}, \frac{7 π}{9}, \frac{11 π}{9}, \frac{13 π}{9}, \frac{17 π}{9}$

$Hence, \sum θ_{i} = 6 π$

Solution

Q.
Let $| \cos θ \cos (60 - θ) \cos (60 + θ) |$ $\leq \frac{1}{8},$ $θ \in [0, 2 π] .$ Then, the sum of all $θ \in [0, 2 π],$ where $\cos 3 θ$ attains its maximum value, is: [2024]

1 $18 π$

2 $9 π$

3 $6 π$

4 $15 π$

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Solution

Q. Let |cosθcos(60-θ)cos(60+θ)|≤18,θ∈[0,2π]. Then, the sum of all θ∈[0,2π], where cos3θ attains its maximum value, is: [2024]

1 18π

2 9π

3 6π

4 15π