Medium MCQ +4 / -1 PYQ · JEE Mains 2022

The number of solutions of the equation

$$\cos \left( {x + {\pi \over 3}} \right)\cos \left( {{\pi \over 3} - x} \right) = {1 \over 4}{\cos ^2}2x$$, $x \in [ - 3\pi ,3\pi ]$ is :

  1. A 8
  2. B 5
  3. C 6
  4. D 7 Correct answer

Solution

<p>$$\cos \left( {x + {\pi \over 3}} \right)\cos \left( {{\pi \over 3} - x} \right) = {1 \over 4}{\cos ^2}2x,\,x \in [ - 3\pi ,\,3\pi ]$$</p> <p>$\Rightarrow \cos 2x + \cos {{2\pi } \over 3} = {1 \over 2}{\cos ^2}2x$</p> <p>$\Rightarrow {\cos ^2}2x - 2\cos 2x - 1 = 0$</p> <p>$\Rightarrow \cos 2x = 1$</p> <p>$\therefore$ $x = - 3\pi ,\, - 2\pi ,\, - \pi ,\,0,\,\pi ,\,2\pi ,\,3\pi$</p> <p>$\therefore$ Number of solutions = 7</p>

About this question

Subject: Mathematics · Chapter: Trigonometric Functions · Topic: Trigonometric Ratios and Identities

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