To solve the equation sin(x)cos(x)cos(2x)cos(4x) = 0, we need to find the values of x that make the equation true.

First, let's consider each factor separately:

sin(x) = 0 when x = 0, π, 2π, ...

cos(x) = 0 when x = π/2, 3π/2, 5π/2, ...

cos(2x) = 0 when 2x = π/2, 3π/2, 5π/2, ...
Solving for x, we get x = π/4, 3π/4, 5π/4, 7π/4, ...

cos(4x) = 0 when 4x = π/2, 3π/2, 5π/2, ...
Solving for x, we get x = π/8, 3π/8, 5π/8, 7π/8, 9π/8, 11π/8, 13π/8, 15π/8, ...

Now, let's find the values of x that satisfy the equation by considering all possible combinations of the values we found for each factor:

1. x = 0, π, 2π, ... (sin(x) = 0)
2. x = π/2, 3π/2, 5π/2, ... (cos(x) = 0)
3. x = π/4, 3π/4, 5π/4, 7π/4, ... (cos(2x) = 0)
4. x = π/8, 3π/8, 5π/8, 7π/8, 9π/8, 11π/8, 13π/8, 15π/8, ... (cos(4x) = 0)

By considering all possible combinations, we find the following values of x that satisfy the equation:

x = π/2, 3π/2, 5π/2, ...
x = π/4, 3π/4, 5π/4, 7π/4, ...
x = π/8, 3π/8, 5π/8, 7π/8, 9π/8, 11π/8, 13π/8, 15π/8, ...

Therefore, the solution to the equation sin(x)cos(x)cos(2x)cos(4x) = 0 is x = π/2 + nπ, π/4 + nπ/2, π/8 + nπ/4, where n is an integer.