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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