3. For $x<0$ there are no solutions, since the function on the right is negative, while the one on the left is non-negative. For $x \geqslant 0$ we have $\arccos \left(\frac{1-x^{2}}{1+x^{2}}\right)=2 \operatorname{arctg} x$, therefore the original inequality is equivalent to

$$
(x-a)^{2} \leqslant 0
$$

From which $x=a$ and for $a \geqslant 0$ there is one solution $x=a$.
Answer: When $a<0$, there are no solutions; when $a \geqslant 0$, there is one solution $x=a$.