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JEE Main 202224 JuneEvening Shift
Exam
JEE Main
Year
2022
Date
24 June
Shift
Evening Shift
Subject
Mathematics
Chapter
Trigonometric Equations

JEE Main 2022 · 24 June · Evening Shift · Mathematics · Trigonometric Equations

The number of solutions of the equation (x + (π)/(3)) ((π)/(3) - x)...

The number of solutions of the equation cos⁡(x+π3)cos⁡(π3−x)=14cos⁡22x\cos \left(x + \dfrac{\pi}{3}\right)\cos \left(\dfrac{\pi}{3} - x\right) = \dfrac{1}{4}\cos^{2}2x, x∈[−3π,3π]x \in \left[- 3\pi, 3\pi\right] is:

Options

  1. A88
  2. B55
  3. C66
  4. D77

Correct answer

D. 77

Step-by-step solution

Given,

cos⁡(x+π3)cos⁡(x−π3)=14cos⁡2(2x)\cos \left(x + \dfrac{\pi}{3}\right)\cos \left(x - \dfrac{\pi}{3}\right) = \dfrac{1}{4}\cos^{2}\left(2x\right)

⇒2cos⁡(x+π3)cos⁡(x−π3)=12cos⁡2(2x)⇒ 2\cos \left(x + \dfrac{\pi}{3}\right)\cos \left(x - \dfrac{\pi}{3}\right) = \dfrac{1}{2}\cos^{2}\left(2x\right)

⇒cos⁡(2x)+cos⁡(2π3)=12cos⁡22x⇒ \cos \left(2x\right) + \cos \left(\dfrac{2\pi}{3}\right) = \dfrac{1}{2}\cos^{2}2x

⇒cos⁡2x+(−12)=12cos⁡22x⇒ \cos 2x + \left(- \dfrac{1}{2}\right) = \dfrac{1}{2}\cos^{2}2x

⇒cos⁡22x−2cos⁡2x+1=0⇒ \cos^{2}2x - 2\cos 2x + 1 = 0

⇒(cos⁡2x−1)2=0⇒ {\left(\cos 2x - 1\right)}^{2} = 0or cos⁡2x=1\cos 2x = 1

⇒2x=−6π,−4π,−2π,0,2π,4π,6π⇒ 2x = - 6\pi, - 4\pi, - 2\pi, 0, 2\pi, 4\pi, 6\pi

So, x∈{−3π,−2π,−π,0,π,2π,3π}x \in \left\{- 3\pi, - 2\pi, - \pi, 0, \pi, 2\pi, 3\pi\right\}

So total 77 solutions.

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