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

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

Substitute these values:

(2cos^2(x) - 1)cos(x) - 2sin(x)cos(x)sin(x) = 0
2cos^3(x) - cos(x) - 2sin(x)cos(x)sin(x) = 0
2cos^3(x) - cos(x) - sin(2x)sin(x) = 0
2cos^3(x) - cos(x) - (2sin(x)cos(x))sin(x) = 0
2cos^3(x) - cos(x) - 2sin^2(x)cos(x) = 0

Now, we have a cubic equation in terms of cos(x). If we assume cos(x) = y, the equation becomes:

2y^3 - y - 2(1 - y^2)y = 0
2y^3 - y - 2y + 2y^3 = 0
4y^3 - 3y = 0
y(4y^2 - 3) = 0
y = 0 or 4y^2 - 3 = 0

If y = 0, then cos(x) = 0, which implies x = π/2 + πk, where k is an integer.

If 4y^2 - 3 = 0:

4y^2 = 3
y^2 = 3/4
y = ±sqrt(3)/2

This implies that cos(x) = ±sqrt(3)/2, which occurs when x = π/6 + 2πk or x = 5π/6 + 2πk, where k is an integer.

Therefore, the solutions to the equation cos(2x)cos(x) - sin(2x)sin(x) = 0 are:
x = π/2 + πk, π/6 + 2πk, 5π/6 + 2πk.