3. Given the equation $x^{2}+(b+2) x y+b y^{2}=0$ $(b \in \mathbf{R})$ represents two lines. Then the range of the angle between them is $\qquad$ .
3. $\left[\arctan \frac{2 \sqrt{5}}{5}, \frac{\pi}{2}\right]$.
When $b=0$, the two lines represented are $x=0, x+2 y=0$, and their angle is $\arctan 2$.
$$
\begin{array}{l}
\text{When } b \neq 0, \text{ we have } b\left(\frac{y}{x}\right)^{2}+(b+2) \frac{y}{x}+1=0, \\
\Delta=(b+2)^{2}-4 b=b^{2}+4>0 .
\end{array}
$$

Let the slopes of the two lines be $k_{1}$ and $k_{2}$, then
$$
k_{1}+k_{2}=-\frac{b+2}{b}, k_{1} k_{2}=\frac{1}{b} .
$$

When $b=-1$, $k_{1} k_{2}=-1$, at this time the angle is $\frac{\pi}{2}$.
When $b \neq-1$, $\tan \theta=\left|\frac{k_{1}-k_{2}}{1+k_{1} k_{2}}\right|=\frac{\sqrt{b^{2}+4}}{|b+1|}$.
Let $t=b+1$, then
$$
\begin{array}{l}
\tan \theta=\frac{\sqrt{(t-1)^{2}+4}}{|t|}=\sqrt{\frac{t^{2}-2 t+5}{t^{2}}} \\
=\sqrt{1-\frac{2}{t}+\frac{5}{t^{2}}}=\sqrt{\frac{4}{5}+5\left(\frac{1}{t}-\frac{1}{5}\right)^{2}} \\
\geqslant \frac{2 \sqrt{5}}{5} .
\end{array}
$$

In summary, the range of the angle is $\left[\arctan \frac{2 \sqrt{5}}{5}, \frac{\pi}{2}\right]$.