We have $x = \frac{[x^2] + 1}{2}$. Hence, $x = t$, or $x = t + \frac{1}{2}$, where $t$ is an integer.

If $x = t$, we have
$$[x^2] - 2x + 1 = [t^2] - 2t + 1 = t^2 - 2t + 1 = (t - 1)^2 = 0 \Leftrightarrow t = 1 \Leftrightarrow x = 1.$$

If $x = t + \frac{1}{2}$, then
$$[x^2] - 2x + 1 = [(t + \frac{1}{2})^2] - 2(t + \frac{1}{2}) + 1 = [t^2 + t + \frac{1}{4}] - 2t - 1 + 1.$$
Since $t^2 + t + \frac{1}{4}$ is not an integer, $[t^2 + t + \frac{1}{4}] = t^2 + t$.
So,
$$[x^2] - 2x + 1 = t^2 + t - 2t - 1 + 1 = t^2 - t.$$
Thus,
$$t^2 - t = 0 \Leftrightarrow t = 0 \text{ or } t = 1 \Leftrightarrow x = \frac{1}{2} \text{ or } x = \frac{3}{2}.$$

Therefore, the solutions are $x = 1$, $x = \frac{1}{2}$, and $x = \frac{3}{2}$.