Example 5 Let the complex number $z$ satisfy $\left|\arg \left(\frac{z+1}{z+2}\right)\right|=\frac{\pi}{6}$, find the range of $\arg z$.

Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
Let $\arg \left(\frac{z+1}{z+2}\right)=\theta$, then $z+1=\lambda(z+2)(\cos \theta + i \sin \theta), \lambda \geqslant 0$.

As shown in Figure 18-2, let $A(-2,0), B(-1,0)$. Then $\overrightarrow{B Z}=\lambda \cdot \overrightarrow{A Z} \cdot (\cos \theta + i \sin \theta)$. When $\theta=\frac{\pi}{6}$, the vector $\overrightarrow{A Z}$ rotates counterclockwise by $\theta$ and becomes parallel to $\overrightarrow{B Z}$, hence
$$
\angle A Z B \equiv \frac{\pi}{6} .
$$

Thus, the trajectory of $z$ is part of a circle with center $O_{1}\left(-\frac{3}{2}, \frac{\sqrt{3}}{2}\right)$ and radius 1, specifically the major arc $\overparen{A B}$ (including $B$, excluding $A$). When $\theta=-\frac{\pi}{6}$, by symmetry, the other part of the trajectory of $z$ is a major arc $\overparen{A B}$ of a circle with center $O_{2}\left(-\frac{3}{2}, -\frac{\sqrt{3}}{2}\right)$ and radius 1 (including $B$, excluding $A$). It is easy to calculate that
$$
\begin{array}{l}
\tan \alpha=\frac{\sqrt{3}}{3}, \alpha=\frac{\pi}{6} ; \\
\sin \beta=\frac{\sqrt{3}}{3}, \beta=\arcsin \frac{\sqrt{3}}{3} .
\end{array}
$$

Therefore, the range of $\arg z$ is
$$
\left(\frac{5 \pi}{6}-\arcsin \frac{\sqrt{3}}{3}, \pi\right) \cup\left(\pi, \frac{7 \pi}{6}+\arcsin \frac{\sqrt{3}}{3}\right) \text {. }
$$

Note: Generally, the argument equation $\arg \left(\frac{z-z_{1}}{z-z_{2}}\right)=\theta$
$$
\Leftrightarrow \overrightarrow{Z_{1} \vec{Z}}=\lambda \cdot \overrightarrow{Z_{2} Z} \cdot (\cos \theta + i \sin \theta), \lambda>0 .
$$

represents the trajectory of a circular arc with $Z_{1} Z_{2}$ as the chord.