$ -1 \le \frac{1}{x^2 - 2x - 2} \le 1 $

$ \Rightarrow 1 + x^2 - 2x - 2 \ge 0 \Rightarrow \frac{(x-1)^2 - 2}{(x-1)^2 - 3} \ge 0 $

$ \Rightarrow \frac{(x-1-\sqrt{2})(x-1+\sqrt{2})}{(x-1-\sqrt{3})(x-1+\sqrt{3})} \ge 0 $

$ \Rightarrow x \in (-\infty, 1-\sqrt{3}] \cup [1-\sqrt{2}, 1+\sqrt{2}] \cup [1+\sqrt{3}, \infty) \quad ...(1)$

$ \Rightarrow 1 - \frac{1}{x^2 - 2x - 2} \ge 0 \Rightarrow \frac{x^2 - 2x - 3}{x^2 - 2x - 2} \ge 0 $

$ \Rightarrow \frac{(x+1)(x-3)}{(x-1-\sqrt{3})(x-1+\sqrt{3})} \ge 0 $

$ \Rightarrow x \in (-\infty, -1] \cup [1-\sqrt{3}, 1+\sqrt{3}] \cup [3, \infty) \quad ...(2)$

$(1) \cap (2)$

$ \Rightarrow x \in (-\infty, -1] \cup [1-\sqrt{2}, 1+\sqrt{2}] \cup [3, \infty) $

$ \therefore \alpha + \beta + \gamma + \delta = 4 $