Let cos θ cos (60-θ) cos (60-θ) leq 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 :

Mathematics · Trigonometry Ratios and Identities

JEE Main 2024 — 9 April, Shift 1 — Question 9

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 :

Option A:
$9 \pi$
Option B:
$18 \pi$
Option C:
$6 \pi$
Correct
Option D:
$15 \pi$

Answer: C

Step-by-step solution

We know that $(\cos \theta)\left(\cos \left(60^{\circ}-\theta\right))\left(\cos \left(60^{\circ}+\theta\right))=\frac{1}{4} \cos 3 \theta\right.\right.$

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 $\frac{\pi}{9}+\frac{5 \pi}{9}+\frac{7 \pi}{9}+\frac{11 \pi}{9}+\frac{13 \pi}{9}+\frac{17 \pi}{9}=\frac{54 \pi}{9}=6 \pi$

Answer key and solution verified before publishing.

Practise Trigonometry Ratios and Identities

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2024
Paper: 9 April, Shift 1
Subject: Mathematics
Chapter: Trigonometry Ratios and Identities
Topic: Periodicity of trigometric functions,Solutions of trigonometric equations