If , then the number of real values of , which satisfy the equation , is
- A.3
- B.5
- C.7
- D.9
Show answer and solution
Correct option: C
Solution
Group and apply sum-to-product:
(cos x + cos 3x) + (cos 2x + cos 4x) = 0
→ 2cos2x·cosx + 2cos3x·cosx = 0
→ 2cosx(cos2x + cos3x) = 0
→ 2cosx · 2cos(5x/2)cos(x/2) = 0
So either cosx = 0, or cos(5x/2) = 0, or cos(x/2) = 0.
cos(x/2) = 0 → x/2 = π/2 → x = π (one solution in [0,2π))
cosx = 0 → x = π/2, 3π/2 (two solutions)
cos(5x/2) = 0 → 5x/2 = π/2, 3π/2, 5π/2, 7π/2, 9π/2 → x = π/5, 3π/5, π, 7π/5, 9π/5 (five solutions, but x = π already counted)
Total distinct values: π/2, 3π/2, π/5, 3π/5, π, 7π/5, 9π/5 → 7 solutions.
(cos x + cos 3x) + (cos 2x + cos 4x) = 0
→ 2cos2x·cosx + 2cos3x·cosx = 0
→ 2cosx(cos2x + cos3x) = 0
→ 2cosx · 2cos(5x/2)cos(x/2) = 0
So either cosx = 0, or cos(5x/2) = 0, or cos(x/2) = 0.
cos(x/2) = 0 → x/2 = π/2 → x = π (one solution in [0,2π))
cosx = 0 → x = π/2, 3π/2 (two solutions)
cos(5x/2) = 0 → 5x/2 = π/2, 3π/2, 5π/2, 7π/2, 9π/2 → x = π/5, 3π/5, π, 7π/5, 9π/5 (five solutions, but x = π already counted)
Total distinct values: π/2, 3π/2, π/5, 3π/5, π, 7π/5, 9π/5 → 7 solutions.
[JEE Main 2016]