Example 4 - For a $4n+2$-sided polygon $A_{1} A_{2} A_{3} \cdots A_{4 n+2}$ (where $n$ is a natural number), each interior angle is an integer multiple of $30^{\circ}$.
Given the quadratic equations in $x$
$$
\begin{array}{l}
x^{2}+2 x \sin A_{1}+\sin A_{2}=0, \\
x^{2}+2 x \sin A_{2}+\sin A_{3}=0, \\
x^{2}+2 x \sin A_{3}+\sin A_{1}=0
\end{array}
$$

all have real roots, find the degree measures of each interior angle of this convex $4n+2$-sided polygon. (85 Fuzhou and other three cities league question)
Since each interior angle is an integer multiple of $30^{\circ}$, the possible angles are $30^{\circ}, 60^{\circ}, 90^{\circ}, 120^{\circ}, 150^{\circ}$, with their respective sine values being $\frac{1}{2}, \frac{\sqrt{3}}{2}, 1$.
If $\sin A_{1}=\frac{1}{2}$, by $\sin A_{2} \geqslant \frac{1}{2}, \sin A_{3} \geqslant \frac{1}{2}$, the discriminant of equation (1) is $\Delta_{1}=4\left(\sin ^{2} A_{1} - \sin A_{2}\right) \leqslant 4\left(\frac{1}{4}-\frac{1}{2}\right)<0$. Therefore, equation (1) has no real roots, which contradicts the given condition, so $\sin A_{1} \neq \frac{1}{2}$.
Similarly, if $\sin A_{1}=\frac{\sqrt{3}}{2}$, equation (1) also has no real roots. Hence, $\sin A_{1} \neq \frac{\sqrt{3}}{2}$. Therefore, $\sin A_{1} = 1$, so $A_{1}=90^{\circ}$. Similarly, $\sin A_{2}=\sin A_{3}=1$, so $A_{2}=A_{3}=90^{\circ}$. Thus, the sum of the remaining $4 n-1$ interior angles is $4 n \times 180^{\circ}-3 \times 90^{\circ}=720^{\circ} n-270^{\circ}$.
Since these angles are all no greater than $150^{\circ}$, we have
$$
720^{\circ} n-270^{\circ} \leqslant (4 n-1) \times 150^{\circ},
$$

then $n \leqslant 1$. Since $n$ is a natural number, $n=1$. Therefore, the convex polygon is a hexagon,
$$
A_{4}+A_{5}+A_{6}=4 \cdot 180^{\circ}-3 \cdot 90^{\circ}=450^{\circ} \text{. }
$$

Since $A_{4}, A_{5}, A_{6} \leqslant 150^{\circ}$, we have
$$
A_{4}=A_{5}=A_{6}=150^{\circ} .
$$