3. Let the function $f(x)$ have the domain $\mathbf{R}$, and when $x \geqslant 0$, $f(x)=|x-2|+a$ (where $a$ is a real number). If $f(x)$ is an odd function, then the solution set of the inequality $f(x) \geqslant 1$ is $\qquad$ .
Given $f(0)=2+a=0 \Rightarrow a=-2$,
then for $x \geqslant 0$, $f(x)=|x-2|-2 \geqslant 1 \Rightarrow|x-2| \geqslant 3 \Rightarrow x \geqslant 5$;
$$
f(x)=|x-2|-2 \leqslant-1 \Rightarrow|x-2| \leqslant 1 \Rightarrow 1 \leqslant x \leqslant 3 .
$$

By symmetry, the solution set for $f(x) \geqslant 1$ is $[-3,-1] \bigcup[5,+\infty)$.