ExamHabit
Practise free

JEE Main 2018

JEE Main 2018 Mathematics question on Trigonometry

If sum of all the solutions of the equation 8cos⁡x⋅(cos⁡(π6+x)⋅cos⁡(π6−x)−12)=18\cos x \cdot \left(\cos\left(\dfrac{\pi}{6} + x\right) \cdot \cos\left(\dfrac{\pi}{6} - x\right) - \dfrac{1}{2}\right) = 1 in [0,π][0, \pi] is kπk\pi, then kk is equal to
  1. A.
    23\dfrac{2}{3}
  2. B.
    139\dfrac{13}{9}
  3. C.
    89\dfrac{8}{9}
  4. D.
    209\dfrac{20}{9}
Show answer and solution

Correct option: B

Solution

cos(π/6 + x)·cos(π/6 - x) = cos²(π/6) - sin²x = 3/4 - sin²x = 3/4 - (1 - cos²x) + ...

More directly: cos(A+B)cos(A-B) = cos²A - sin²x = 3/4 - sin²x.

So the bracket becomes: (3/4 - sin²x) - 1/2 = 1/4 - sin²x = 1/4 - (1 - cos²x) = cos²x - 3/4.

The equation becomes: 8cosx·(cos²x - 3/4) = 1
→ 8cos³x - 6cosx = 1
→ 2(4cos³x - 3cosx) = 1... wait: 8cos³x - 6cosx = 2(4cos³x - 3cosx) = 2cos(3x).

So: 2cos(3x) = 1 → cos(3x) = 1/2.

For x ∈ [0, π], 3x ∈ [0, 3π].
cos(3x) = 1/2 gives 3x = π/3, 5π/3, 7π/3.
So x = π/9, 5π/9, 7π/9.

Sum = (π/9)(1 + 5 + 7) = 13π/9, so k = 13/9.

[JEE Main 2018]

Practise Trigonometry the way it is asked

A timed set of real JEE Main Trigonometry questions, graded as the exam grades. No sign-up.

Start a Trigonometry test

More: all JEE Main Trigonometry questions · JEE Main 2018 · Mathematics chapter list