For , the equation has
- A.infinitely many solutions
- B.three solutions
- C.one solution
- D.no solution
Show answer and solution
Correct option: D
Solution
The left side can be rewritten. Note sin x + 2sin2x - sin3x = sin x - sin3x + 2sin2x = -2cos2x·sinx + 2sin2x = 2sin x(1 - cos x) ... let me recompute carefully.
sin x - sin3x = -2cos2x·sinx, and 2sin2x = 4sinx·cosx.
So LHS = -2cos2x·sinx + 4sinx·cosx = 2sinx(-cos2x + 2cosx) = 2sinx(-(2cos²x-1) + 2cosx) = 2sinx(1 + 2cosx - 2cos²x).
For x ∈ (0,π), sinx > 0, so we need 2sinx(1 + 2cosx - 2cos²x) = 3.
The maximum of sinx is 1, so we need the factor (1 + 2cosx - 2cos²x) to be large enough. Let t = cosx ∈ (-1,1). The expression 1 + 2t - 2t² has maximum at t = 1/2: value = 1 + 1 - 1/2 = 3/2. So the maximum of the entire LHS is at most 2·1·(3/2) = 3, but sinx = 1 and cosx = 1/2 cannot hold simultaneously (sinx=1 requires x=π/2, cosx=1/2 requires x=π/3). At x=π/2: cosx=0, LHS
sin x - sin3x = -2cos2x·sinx, and 2sin2x = 4sinx·cosx.
So LHS = -2cos2x·sinx + 4sinx·cosx = 2sinx(-cos2x + 2cosx) = 2sinx(-(2cos²x-1) + 2cosx) = 2sinx(1 + 2cosx - 2cos²x).
For x ∈ (0,π), sinx > 0, so we need 2sinx(1 + 2cosx - 2cos²x) = 3.
The maximum of sinx is 1, so we need the factor (1 + 2cosx - 2cos²x) to be large enough. Let t = cosx ∈ (-1,1). The expression 1 + 2t - 2t² has maximum at t = 1/2: value = 1 + 1 - 1/2 = 3/2. So the maximum of the entire LHS is at most 2·1·(3/2) = 3, but sinx = 1 and cosx = 1/2 cannot hold simultaneously (sinx=1 requires x=π/2, cosx=1/2 requires x=π/3). At x=π/2: cosx=0, LHS
[JEE Advanced 2014]