The domain of $\sin^{-1}(x)$ is $-1 \le x \le 1$. Therefore, for $y = \sin^{-1}(x^2 - 4)$ to be defined, we must have:

$-1 \le x^2 - 4 \le 1$

We can split this into two inequalities:

1. $x^2 - 4 \le 1 \Rightarrow x^2 \le 5 \Rightarrow -\sqrt{5} \le x \le \sqrt{5}$

2. $x^2 - 4 \ge -1 \Rightarrow x^2 \ge 3 \Rightarrow x \le -\sqrt{3}$ or $x \ge \sqrt{3}$

Combining these two conditions, we have:

$-\sqrt{5} \le x \le -\sqrt{3}$ or $\sqrt{3} \le x \le \sqrt{5}$