Answer :

Given equation is


cos 4x = 1 + k sin2x cos2x


Now consider the LHS of the equation,


cos 4x = 2cos2 2x – 1


[Formula for Cos 2x = 2cos2 x – 1]


= 2[2cos2 x - 1]2 – 1


= 2[(2cos2 x)2 – 2 × (2 cos2 x) × (1) + (1)2] – 1


[Applying (a-b)2 = a2 – 2ab + b2 formula]


= 2[4cos4 x – 4cos2 x +1] -1


= 8 cos4 x – 8cos2 x +2 – 1


= 8cos2 x (cos2 x – 1) + 1


= 8cos2 x (-sin2 x) +1


= - 8cos2 x sin2 x + 1


Now as per the LHS cos 4x = - 8cos2 x sin2 x + 1 -------- (1)


Comparing LHS with the RHS,


cos 4x = 1 - 8cos2 x sin2 x = 1 + k sin2x cos2x


by comparing we get k = -8


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