Question

The number of solutions of the equation cosx = sin2x, x ∈ [0, 6π] is

Answer 1

Given : cosx = sin2x, x ∈ [0, 6π]  

To find : number of solutions

Solution:

Cosx = Sin2x

=> Cosx = 2SinxCosx

=> Cosx - 2SinxCosx = 0

=> Cosx(1  - 2Sinx) =  0

=> Cosx = 0   1 - 2Sinx = 0

Cosx = 0 =>   x  = π/2 ,  3π/2  , 5π/2 , 7π/2 , 9π/2 , 11π/2

1 - 2Sinx = 0 =>  Sinx = 1/2

=> x = π/6 ,  5π/6  ,   13π/6 ,  17π/6  ,   25π/6 ,  29π/6  ,  

π/6 ,  π/2 ,  5π/6 , 3π/2  , 13π/6 , 5π/2 , 17π/6 , 7π/2 ,  25π/6 , 9π/2 , 29π/6 , 11π/2

Number of Solutions = 12  

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