To solve the equation 2sin(2x) - sin(x)cos(x)cos(2x) = 0, we can rewrite sin(2x) and cos(2x) in terms of sin(x) and cos(x):

sin(2x) = 2sin(x)cos(x)
cos(2x) = cos^2(x) - sin^2(x)

Substitute these values into the equation:

2(2sin(x)cos(x)) - sin(x)cos(x)(cos^2(x) - sin^2(x)) = 0
4sin(x)cos(x) - sin(x)cos(x)(cos^2(x) - sin^2(x)) = 0
4sin(x)cos(x) - sin(x)cos(x)cos^2(x) + sin(x)cos(x)sin^2(x) = 0
4sin(x)cos(x) - cos(x)sin(x)cos^2(x) + sin^2(x)cos(x) = 0
4sin(x)cos(x) - cos^2(x)sin(x) + sin^2(x)cos(x) = 0

Now, we can factor out a cos(x) from the equation:

cos(x)(4sin(x) - cos(x)sin(x) + sin^2(x)) = 0
cos(x)(4sin(x) - sin^2(x)) = 0
cos(x)(sin(x)(4 - sin(x))) = 0

Now, we set each factor equal to zero:

cos(x) = 0
sin(x)(4 - sin(x)) = 0

From the first equation, cos(x) = 0 when x = π/2 + nπ, where n is an integer.

From the second equation, sin(x) = 0 or 4 - sin(x) = 0:

sin(x) = 0 when x = nπ, where n is an integer.4 - sin(x) = 0
sin(x) = 4
However, sin(x) cannot be greater than 1, so this solution is not valid.

Therefore, the solutions to the equation are x = nπ, x = π/2 + nπ, where n is an integer.