Let substitute: $x_n = 2 \cos t_n$, where $t_n \in [0,\pi]$. Hence $\cos t_{n+1} = \cos 2t_n \, \Rightarrow \, x_n = 2\cos (2^{n-1} t_1)$. Let assume that $t_1 \neq 0,\, t_1 \neq \pi$. These two cases can be considered apart. Thus we have: $ x_1x_2\ldots x_n = 2^n \cos t_1 \cos 2t_1 \ldots \cos 2^{n-1} t_1 = \frac{\sin 2^n t_1}{\sin t_1}$. Because $\sin t_1 > 0 $ when $t_n \in (0,\pi)$, the sign of $ x_1x_2\ldots x_n $ is equal to the sign of $\sin 2^n t_1 $ and this sign is equal to: $ (-1)^{\lfloor \frac{2^n t_1}{\pi} \rfloor} $. Let denote $s_1 = \frac{t_1}{\pi}$. Obviously $s_1 \in (0,1)$. Consider the decimal representation of $s_1$ and let it be: $0.a_1a_2 \ldots a_n, \ldots$, where $a_i\in \{0,1\}$. Notice that $\lfloor 2^n s_1 \rfloor $ is odd if and only if $a_n =1$, so $A= s_1$. Finally $ A =\frac{1}{\pi}\cos^{-1}\left(\frac{x_{1}}{2}\right) $, where $\cos^{-1}$ means $\arccos$.