We can start by using the double angle formula for sine and cosine:

sin 2x = 2 sin x cos x
cos 2x = cos^2 x - sin^2 x = 2 cos^2 x - 1 = 1 - 2 sin^2 x

Using these formulas, we can rewrite the given equation as:

sin 4x cos 4x = sin x cos x
2 sin 2x cos 2x = sin x cos x
2 (2 sin x cos x) (1 - 2 sin^2 x) = sin x cos x
4 sin x cos x - 8 sin^3 x cos x = sin x cos x
3 sin x cos x = 8 sin^3 x cos x
3 = 8 sin^2 x
sin x = ±√(3/8)

Since we are looking for solutions in the interval [0, 2π], we need to consider the solutions in each quadrant separately. In the first quadrant, both sin x and cos x are positive, so we get:

sin x = √(3/8)
cos x = √(1 - 3/8) = √(5/8)

In the second quadrant, sin x is positive and cos x is negative, so we get:

sin x = √(3/8)
cos x = -√(5/8)

In the third quadrant, both sin x and cos x are negative, so we get:

sin x = -√(3/8)
cos x = -√(5/8)

In the fourth quadrant, sin x is negative and cos x is positive, so we get:

sin x = -√(3/8)
cos x = √(5/8)

Therefore, there are four solutions in the interval [0, 2π]:

x = arctan(√(3/5)), arctan(-√(3/5)), arctan(-√(3/5)) + π, arctan(√(3/5)) + π

Note that the solutions in the third and fourth quadrants are obtained by adding π to the solutions in the first and second quadrants, respectively, due to the periodicity of the trigonometric functions.