1. The solution set of the inequality $(x-1) \sqrt{x^{2}-x-2} \geqslant 0$ is $\qquad$ .
2. $\{-1 \mid \cup\{x \mid x \geqslant 2\}$.
Solution 1: Classify by the sign of $x-1$.
(1) When $x-1 \geqslant 0$, it must be that
$$
\sqrt{x^{2}-x-2} \geqslant 0 \text {. }
$$

That is, $\left\{\begin{array}{l}x \geqslant 1, \\ x \leqslant-1\end{array}\right.$ or $x \geqslant 2$.
Thus, $x \geqslant 2$.
(2) When $x-10$, it must be that
$$
x-1 \geqslant 0 \text {. }
$$

That is, $\left\{\begin{array}{l}x^{2}-x-2>0, \\ x \geqslant 1 .\end{array}\right.$
Thus, $x>2$.
(2) When $\sqrt{x^{2}}-x-2=0$, $x-1$ can take any value, giving
$x=-1$ or $x=2$.
Combining (3) and (4), we get $\{-1\} \cup|x| x \geqslant 2\}$.
Solution 3: Use the complement method. First, the domain of the inequality is
$x \leqslant-1$ or $x \geqslant 2$.
Second, solve the inequality $(x-1) \sqrt{x^{2}-x-2}0 .\end{array}\right.$

Thus,
In (5), find the complement of (6) to get
$\left\{\begin{array}{l}x \leqslant-1 \text { or } x \geqslant 2, \\ x \geqslant-1 .\end{array}\right.$

Thus, $x=-1$ or $x \geqslant 2$.