JEE Advanced | 2014Trigonometric EquationMedium
Question
For x ∈ (0, π) the equation sinx + 2sin2x - sin3x = 3 has
Options
A.Infinitely many solutions
B.Three solutions
C.One solution
D.No solution
Solution
sinx + 2sin2x - sin3x = 3
2cos2x(-sinx) + 4sinxcosx = 3
2sinx[2cosx - cos2x] = 3
2sinx(2cosx - (2cos2x - 1)) = 3
2sinx(1 + 2cosx - 2cos2x) = 3
2 sin x
= 3
Possible only when
sinx = 1 .....(i)
and
= 0
⇒ cos x =
.....(ii)
From (i) and (ii)
No solution.
2cos2x(-sinx) + 4sinxcosx = 3
2sinx[2cosx - cos2x] = 3
2sinx(2cosx - (2cos2x - 1)) = 3
2sinx(1 + 2cosx - 2cos2x) = 3
2 sin x
= 3Possible only when
sinx = 1 .....(i)
and
= 0⇒ cos x =
.....(ii)From (i) and (ii)
No solution.
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