JEE Main 2022 · 24 June · evening · Single correct · +4 / −1

The number of solutions of the equation
cos ( x + π/3 )cos ( π/3 - x ) = 1/4cos ²2x, x ∈ [ - 3π ,3π ] is :

  1. (a) 8
  2. (b) 5
  3. (c) 6
  4. (d) 7
Answer and solution

Answer (d) 7

Solution

Maths sir · 4 steps

The left side is a cos(A+B)cos(A−B) product, so turn it into cos²A − sin²B right away.

  1. Use the identity. cos(A+B)cos(A-B) = cos² A - sin² B. With A = π/3 and B = x, the left side becomes 1/4 - sin² x.
  2. Write in cos 2x. Put sin² x = (1 - cos 2x)/2, multiply the whole equation by 4 and collect terms. It becomes a perfect square.

    cos² 2x - 2 cos 2x + 1 = 0 ⇒ (cos 2x - 1)² = 0 ⇒ cos 2x = 1

  3. General solution. cos 2x = 1 gives 2x = 2nπ, so x = nπ with n an integer.
  4. So. In [-3π, 3π] the allowed values are n = -3, -2, -1, 0, 1, 2, 3. That is 7 solutions, option (d).

The idea to keep. Whenever you see cos(A+B)cos(A-B), replace it with cos² A - sin² B. Then write everything in one function, and when you count solutions, check whether the endpoints are included.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Trigonometric Equations

Asked about 50 times in JEE Main since 2020. This question is from 2022.