1.

The no of roots of the equation Sinx+2sin2x=3+sin3x

Answer»

we have to find the no of roots of the EQUATION,

sinx + 2sin2x = 3 + sin3x

solution : sinx + 2sin2x = 3 + sin3x

⇒sinx + 2sin2x = 3 + 3sinx - 4sin³x

[ we know, sin3A = 3sinA - 4sin³A ]

⇒- 2sinx + 2sin2x + 4sin³x = 3

⇒-2sinx (1 - 2sin²x) + 2sin2x = 3

using formula, COS2A = 1 - 2sin²A

and sin2A = 2SINA cosA

⇒-2sinx (cos2x) + 2(2sinx cosx) = 3

⇒-2sinx(cos2x - 2cosx) = 3

⇒2sinx (2cosx - cos2x) = 3

here 2sinx(2cosx - cos2x) takes maximum value at X = π/3

let f(x) = 2sinx(2cosx - 2cos2x)

f(π/3) = 2 × √3/2 (2 × 1/2 + 2 × 1/2) = 3√3/2

here f(π/3) = 3√3/2 < 3

it means, y = 3 never cuts at any point of graph of f(x) as you can see in figure.

hence it is clear that there is no root possible of equation, sinx + 2sin2x = 3 + sin3x



Discussion

No Comment Found

Related InterviewSolutions