I suggest you to transform the difference of like trigonometric functions into a product such that:

sin alpha - sin beta = 2 cos ((alpha+beta)/2)*(sin(alpha-beta)/2)

sin 4x - sin2x = 2 cos ((4x+2x)/2)*(sin(4x-2x)/2)

sin 4x - sin2x = 2 cos (3x)*sin(-x)

sin 4x - sin2x = - 2 cos (3x)*sin(x)

Replacing the difference in equation sin 4x - sin2x = 0 yields:

- 2 cos (3x)*sin(x) = 0 => cos 3x = 0 and sin x = 0

cos 3x = 0 => 3x = pi/2 + 2n*pi => x = pi/6 + 2n*(pi/3)

sin x = 0 => x = 0

The solutions to trigonometric equation are x=0; x=pi/6 + 2n*(pi/3).

We'll recall the double angle identity, such as sin 4x = sin 2*(2x)

But, the formula for the double angle sin 2a = 2sina*cosa.

We'll use the formula to sin 2*(2x):

sin 2*(2x) = 2 sin 2x* cos 2x (1)

Now, we'll re-write the equation, replacing sin 4x by the formula (1):

sin 4x - sin 2x = 0

2 sin 2x* cos 2x - sin 2x = 0

We'll factorize by sin 2x:

sin 2x*(2cos 2x - 1) = 0

We'll cancel each factor:

sin 2x = 0

2x = +/-arcsin 0

2x = 0

x = 0

We'll put 2cos 2x - 1 equal to 0:

2cos 2x - 1 = 0

We'll add 1 both sides:

2cos 2x = 1

We'll divide by 2:

cos 2x = 1/2

2x = +/- arccos (1/2)

2x = +/- (pi/3)

We'll divide by 2:

x = +/- pi/6

The solution -pi/6 could be expressed as x = 2pi - pi/6

x = 11pi/6

All requested solutions of the equation, over the set [0,2pi], are:{0 ; pi/6, 11pi/6}.