sin^4cos^2x=(1/16)(1-cos 2x)(1-cos4x)

Question

work one 1 side only!

Answer 1

pick x=50º
LS = .202..
RS = .379..

Your equation is NOT an identity

I assumed your LS = sin^x4cos^2x
you had sin^2[cos^2x], but I doubt that is what you meant.
(that would be even worse)