Prove the following: sin2x + 2sin4x + sin6x = 4cos2xsin4x

LHS = sin2x + 2sin4x + sin6x
= sin2x + sin6x + 2sin4x
= 2sin(2x+6x/2)cos(2x-6x/2) + 2sin4x
= 2sin4xcos(-2x) + 2sin4x
= 2sin4x[cos(-2x) + 1]
= 2sin4x[cos2x + 1] (Since cos(-x) = cos(x))
= 2sin4x(2cos²x) (Since 1+cos2x = 2cos²x)
= 4cos²xsin4x
This is not equal to 4cos2xsin4x. There appears to be an error in the original question or solution provided. The provided solution is incomplete and uses the identity 1+cos2x=2cos²x incorrectly in the context of the problem.

Let's try another approach assuming the question is correct:
LHS = sin2x + 2sin4x + sin6x
= sin2x + sin6x + 2sin4x
= 2sin(4x)cos(2x) + 2sin4x
= 2sin4x(cos2x + 1)
= 2sin4x(2cos²x)
= 4sin4xcos²x ≠ 4cos2xsin4x

To prove the given equation, we need a different approach or correction in the given equation.