3. Given that the function $f(x)$ is an increasing function defined on $[-4,+\infty)$, to ensure that for all real numbers $x$ in the domain, the inequality $f\left(\cos x-b^{2}\right) \geqslant f\left(\sin ^{2} x-b-3\right)$ always holds, the range of the real number $b$ is $\qquad$.
$$
\left\{\begin{array} { l } 
{ \operatorname { cos } x - b ^ { 2 } \geqslant - 4 , } \\
{ \operatorname { sin } ^ { 2 } x - b - 3 \geqslant - 4 , } \\
{ \operatorname { cos } x - b ^ { 2 } \geqslant \operatorname { sin } ^ { 2 } x - b - 3 }
\end{array} \Rightarrow \left\{\begin{array} { l } 
{ b ^ { 2 } \leqslant \operatorname { cos } x + 4 , } \\
{ b \leqslant \operatorname { sin } ^ { 2 } x + 1 , } \\
{ b ^ { 2 } - b \leqslant \operatorname { cos } x - \operatorname { sin } ^ { 2 } x + 3 }
\end{array} \Rightarrow \left\{\begin{array}{l}
b^{2} \leqslant 3, \\
b \leqslant 1, \\
b^{2}-b \leqslant \frac{7}{4}
\end{array} \Rightarrow \frac{1}{2}-\sqrt{2} \leqslant b \leqslant 1,\right.\right.\right.
$$

So the range of the real number $b$ is $\left[\frac{1}{2}-\sqrt{2}, 1\right]$.