Example 1 Given $A=\left\{x \left\lvert\, \frac{2 x+1}{x-3} \geqslant 1\right.\right\}$, $B=\left\{y \mid y=b \arctan t,-1 \leqslant t \leqslant \frac{\sqrt{3}}{3}, b \leqslant 0\right\}$,
and $A \cap B=\varnothing$.
Find the range of real number $b$.
(2010, Nanjing University Characteristic Test)
Notice that,
$$
\begin{array}{l}
\frac{2 x+1}{x-3} \geqslant 1 \Leftrightarrow \frac{x+4}{x-3} \geqslant 0 \\
\Leftrightarrow x>3 \text { or } x \leqslant-4 .
\end{array}
$$

Therefore, $A=(-\infty,-4] \cup(3,+\infty)$.
From $-1 \leqslant t \leqslant \frac{\sqrt{3}}{3} \Leftrightarrow-\frac{\pi}{4} \leqslant \arctan t \leqslant \frac{\pi}{6}$.
When $b<0$,
$$
B=\left(-\infty, \frac{b \pi}{6}\right] \cup\left[-\frac{b \pi}{4},+\infty\right) \text {. }
$$

At this time, $A \cap B=\varnothing$.
When $b=0$, $B=\{0\}$, at this time, $A \cap B=\varnothing$.
In summary, $b=0$.