To prove this trigonometric identity, we will use double angle formulas and the Pythagorean identity.

Starting with the left-hand side, (1 + cos(4x))sin(2x), we expand the expression using the double angle formula for sin(2x) and the double angle formula for cos(2x):
(1 + cos(4x))(2sin(x)cos(x))
= 2sin(x)cos(x) + 2sin(x)cos(x)cos(4x)
= 2sin(x)cos(x) + 2sin(x)cos(x)(2cos^2(2x) - 1)
= 2sin(x)cos(x) + 4sin(x)cos(x)cos^2(2x) - 2sin(x)cos(x)
= 4sin(x)cos(x)cos^2(2x)

Now, we simplify the right-hand side, cos^2(2x):

Using the double angle formula for cos(2x), we get:
cos^2(2x) = (cos(2x))^2
= (2cos^2(x) - 1)^2
= 4cos^4(x) - 4cos^2(x) + 1

Then, we simplify this further:
4cos^4(x) - 4cos^2(x) + 1 = 4(1 - sin^2(x))^2 - 4(1 - sin^2(x)) + 1
= 4 - 8sin^2(x) + 4sin^4(x) - 4 + 4sin^2(x) + 1
= 4sin^4(x) - 4sin^2(x) + 1

Therefore, (1 + cos(4x))sin(2x) = 4sin(x)cos(x)cos^2(2x) = cos^2(2x), thus proving the trigonometric identity.