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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

JEE Main · Mathematics · Trigonometric Ratios & Identities

The sum of the solutions in (0,2 π) the equation x ((π)/(3)-x)...

The sum of the solutions in (0,2π)(0,2 \pi) the equation cos⁡xcos⁡\cos x \cos\left(\frac{\pi}{3}-x\right) \cos (π3+x)=14\left(\frac{\pi}{3}+x\right)=\frac{1}{4} is

Options

  1. A4π4 \pi
  2. Bπ\pi
  3. C2π2 \pi
  4. D3π3 \pi

Correct answer

C. 2π2 \pi

Step-by-step solution

We know, \cos x \cos \left(\frac{\pi}{3}-x\right) \cos \left(\frac{\pi}{3}+x\right)=14cos⁡\frac{1}{4} \cos3 θ\thetaBut it is given cos⁡xcos⁡(π3−x)cos⁡(π3+x)=14∴cos⁡3θ=1⇒3θ=2π,4π⇒θ=2π3,4π3\begin{aligned}& \cos x \cos \left(\frac{\pi}{3}-x\right) \cos \left(\frac{\pi}{3}+x\right)=\frac{1}{4}\\ \therefore \quad \cos 3 \theta& =1 \\ \Rightarrow \quad 3 \theta& =2 \pi,4 \pi\\ \Rightarrow \quad \theta& =\frac{2 \pi}{3}, \frac{4 \pi}{3} \end{aligned}\thereforeSumofsolutionsSum of solutions=2π3+4π3=6π3=2π\begin{aligned}& =\frac{2 \pi}{3}+\frac{4 \pi}{3}=\frac{6 \pi}{3}\\ & =2 \pi \end{aligned}$
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