Solution:
Squaring the equation
$$\sqrt{\left(4 a^{2}-4 a-1\right) x^{2}-2 a x+1}=1-a x-x^{2}$$
gives the equation
$$x^{2}\left(x^{2}+2 a x-3 a^{2}+4 a-1\right)=0$$
with roots $x_{1}=0$, $x_{2}=1-3 a$ and $x_{3}=a-1$. It is clear that $x_{1}=0$ is a root of (1) for any $a$. On the other hand, $x_{2}=1-3 a$ is a root of (1) if its right-hand side is non-negative, i.e., if
$$1-a(1-3 a)-(1-3 a)^{2} \geq 0 \Longleftrightarrow 5 a-6 a^{2} \geq 0 \Longleftrightarrow a \in\left[0, \frac{5}{6}\right]$$
Analogously, $x_{3}=a-1$ is a root of (1) for $a \in\left[0, \frac{3}{2}\right]$. Two cases are possible.

Case 1. Some of the numbers $x_{1}$, $x_{2}$ and $x_{3}$ are equal. This implies that $a=\frac{1}{3}$, $\frac{1}{2}$ or $1$. It follows from above that $a=\frac{1}{3}$ and $a=\frac{1}{2}$ are solutions of the problem.

Case 2. The numbers $x_{1}$, $x_{2}$ and $x_{3}$ are pairwise different. Then it is easy to see that $a \in\left(\frac{5}{6}, \frac{3}{2}\right] \setminus\{1\}$.

So, the desired values of $a$ are $a=\frac{1}{3}$, $a=\frac{1}{2}$ and $a \in\left(\frac{5}{6}, \frac{3}{2}\right] \setminus\{1\}$.