I need to find the domain of $$\sin^\text{-1}\left(\dfrac{1}{|x^2-1|}\right)$$ Hence, $$-1 \le \dfrac{1}{|x^2-1|} \le 1 \\ \implies \dfrac{1-|x^2-1|}{|x^2-1|} \le 0 $$
Then I made two cases, first that $x^2-1\ge 0$ or $ -1\ge x\ge 1$ and second that, $-1\le x\le 1$
Skipping to the result, $x \in (-\infty, -\sqrt2) \cup (\sqrt2,\infty)$ However, $x=0$ also satisfies this inequality. So, am I forgetting something?
Is it just a coincidence here or should I check all the critical points everytime?
Or if not, then when?
$$\frac{1}{|x^2-1|} \le 1 \implies |x^2-1| \ge 1 \implies x^2-1 \ge 1 ....(1)~or~ x^2-1\le -1...(2).$$ From (1), we get: $x^2 \ge 2 \implies x \le -\sqrt{2}~or~ x \ge \sqrt{2}$.....(3)
From (2), we get $x^2 \le 0 \implies x=0 $......(4)
The final solution is $x\in (-\infty,-\sqrt{2}] \cup [\sqrt{2}, \infty) \cup \{0\}$
OP's way You can only write $$\frac{1}{|x^2-1|} \le 1 \implies \frac{1-|x^2-1|}{|x^2-1|} \le 0$$ there are two cases $$Case (1):~ \text{when}~ x^2 > 1 \implies x <- 1~or~ x>1....(A)$$ Then we get: $$\frac{x^2-2}{x^2-1}\ge 0 \implies (x^-2)(x^2-1) \ge 0 \implies (x+\sqrt{2})(x-\sqrt{2})(x+1)(x-1) \ge 0 \implies x \in (-\infty, -\sqrt{2}) \cup (-1,1) \cup [\sqrt{2}, \infty)....(B)$$ In this case (1), the solution is the overlap of (A) and (B) which is $$x \in (-\infty, -\sqrt{2}] \cup [\sqrt{2}, \infty)....(C)$$
$$Case (2): x^2 <1 \implies -1<x <1....(D)$$ $$\implies \frac{x^2}{1-x^2}\le 0 \implies x^2(x^2-1) \ge 0 \implies (x+1) x^2 (x-1) \ge 0 \implies x=0, x\in (-\infty, -1) \cup (1, \infty).....(E)$$ The overlap of (D) and (E) is just $x=0$
Finally, the solution comes from (C) where $\{0\}$ has to be included.