To solve this equation, we can square both sides:

(x^2-2x+1) = (x-1)^4

Expanding the right side using the binomial theorem:

(x^2-2x+1) = x^4 - 4x^3 + 6x^2 - 4x + 1

Simplifying and rearranging:

x^4 - 4x^3 + 5x^2 - 2x = 0

Factoring out x:

x(x^3 - 4x^2 + 5x - 2) = 0

Using synthetic division or long division, we can find that the cubic factor has a root of x=1:

(x-1)(x^2 - 3x + 2) = 0

So the solutions to the original equation are x=1 and the roots of the quadratic factor:

x^2 - 3x + 2 = 0

Factoring:

(x-1)(x-2) = 0

So the complete set of solutions is x=1, x=2, and x=3.