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The function $g(x) = \cos^{-1}(x^2 - 1)$ is defined when the argument lies in the interval $[-1, 1]$. Thus, we solve: $$-1 \leq x^2 - 1 \leq 1$$ Adding 1 to all parts: $$0 \leq x^2 \leq 2$$ Taking the square root: $$|x| \leq \sqrt{2} \implies -\sqrt{2} \leq x \leq \sqrt{2}$$ So, the domain is $[-\sqrt{2}, \sqrt{2}]$.
Given $\cos^{-1}(x^2 - 1) = \frac{\pi}{3}$, we take the cosine of both sides: $$x^2 - 1 = \cos\left(\frac{\pi}{3}\right)$$ $$x^2 - 1 = \frac{1}{2}$$ $$x^2 = \frac{3}{2} \implies x = \pm\sqrt{\frac{3}{2}} = \pm\frac{\sqrt{6}}{2}$$
The principal branch of $\cos^{-1}x$ is $[0, \pi]$. Other branches are of the form $[2n\pi, (2n+1)\pi]$ for any integer $n \neq 0$. For example, the branch for $n=1$ is $[2\pi, 3\pi]$.
Final Answer: Domain: [-\sqrt{2}, \sqrt{2}], x = \pm\sqrt{1.5}, Range: [2n\pi, (2n+1)\pi], n \in \mathbb{Z}, n \neq 0