Answer
$$[-1,1-\sqrt3)\cup(1+\sqrt3,3]$$
Work Step by Step
$$\ln{(x^{2}-2x-2)}\leq0\\
e^{\ln{(x^{2}-2x-2)}}\leq e^0\\
x^{2}-2x-2\leq1\\ x^{2}-2x-3\leq0\\
(x-3)(x+1)\leq0\\
x\leq3\qquad x\geq-1\\
[-1,3]$$
$$\mbox{Also, }x^{2}-2x-2\gt0\mbox{, as }\ln\mbox{ can only be taken of a positive number.}\\
x^{2}-2x-2\gt0\\
(x-(1+\sqrt 3))(x-(1-\sqrt 3))\gt0\\
(x-1-\sqrt 3)(x-1+\sqrt 3)\gt0\\
x\gt1+\sqrt3\qquad x\lt1-\sqrt3\\
(-\infty,1-\sqrt3)\cup(1+\sqrt3,\infty)$$
$$\mbox{Therefore, the solution is }[-1,1-\sqrt3)\cup(1+\sqrt3,3]$$
