The equation given below has equal roots: What is the value of ' '?

Given $ax^2 + \sin^{-1}(x^2 - 2x + 2) + \cos^{-1}(x^2 - 2x + 2) = 0$. Since $\sin^{-1}u + \cos^{-1}u = \frac{\pi}{2}$ for all $u \in [-1, 1]$, the equation becomes: $ax^2 + \frac{\pi}{2} = 0$. Now consider the argument $u = x^2 - 2x + 2 = (x - 1)^2 + 1$. Since $(x - 1)^2 \ge 0$, the minimum value of $u$ is $1$, occurring at $x = 1$. For $u$ to lie in the domain $[-1, 1]$, we must have $u \le 1$, which requires $(x - 1)^2 + 1 \le 1 \Rightarrow (x - 1)^2 = 0 \Rightarrow x = 1$. Thus, the only real root is $x = 1$. Since the equation has equal roots at $x = 1$: $a(1)^2 + \frac{\pi}{2} = 0 \Rightarrow a + \frac{\pi}{2} = 0 \Rightarrow a = -\frac{\pi}{2}$.