Example 7 Let the complex number $z$ satisfy $\left|\arg \left(\frac{z+1}{z+2}\right)\right|=\frac{\pi}{6}$, find the range of $\arg z$.

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Analysis: The combination of numbers and shapes is an effective method for dealing with problems involving the principal value of an argument.
Solution: Let $\arg \left(\frac{z+1}{z+2}\right)=\theta$, then $z+1=\lambda(z+2) e^{i \theta}, \lambda \geqslant 0$.
As shown in Figure 2-5, let $A(-2,0), B(-1,0)$, then $\overrightarrow{B Z}=\lambda \cdot \overrightarrow{A Z} \cdot e^{i \theta}$.
When $\theta=\frac{\pi}{6}$, the vector $\overrightarrow{A Z}$ rotates counterclockwise by $\theta$ and becomes parallel to $\overrightarrow{B Z}$,

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

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