Given a convex $(4n+2)$-sided polygon $A_{1} A_{2} \cdots A_{4n+2}$ (where $n$ is a positive integer) with each interior angle being an integer multiple of $30^{\circ}$. Consider the 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 the interior angles of this convex $(4n+2)$-sided polygon.
Analyzing according to the given conditions, the degree measures of the internal angles can only be $30^{\circ}, 60^{\circ}, 90^{\circ}, 120^{\circ}$, or $150^{\circ}$, and their sine values can only be $\frac{1}{2}, \frac{\sqrt{3}}{2}$, or 1.
$$
\begin{array}{l}
\Delta=4 \sin ^{2} A_{1}-4 \sin A_{2} \\
\leqslant 4 \sin ^{2} A_{1}-4 \times \frac{1}{2}<0, \\
\end{array}
$$

Then (1) has no real roots, which is a contradiction.
$\therefore \sin A_{1}$ can only take $\frac{\sqrt{3}}{2}$ or 1.
Similarly, $\sin A_{2}$ and $\sin A_{3}$ can only take $\frac{\sqrt{3}}{2}$ or 1.
(2) When $\sin A_{2}=\frac{\sqrt{3}}{2}$, the discriminant of equation (1) is
$$
\begin{aligned}
\Delta & =4 \sin ^{2} A_{1}-4 \sin A_{2} \\
& \leqslant 4 \sin ^{2} A_{1}-4 \times \frac{\sqrt{3}}{2}<0,
\end{aligned}
$$

Then (1) has no real roots, which is a contradiction.
$\therefore \sin A_{1}$ can only take 1.
Similarly, $\sin A_{2}$ and $\sin A_{3}$ can only take 1.
Thus, $\angle A_{1}=\angle A_{2}=\angle A_{3}=90^{\circ}$, and the sum of the remaining $4 n-1$ internal angles is
$$
\begin{array}{l}
\angle A_{4}+\angle A_{5}+\cdots+\angle A_{4+2} \\
=4 n \cdot 180^{\circ}-3 \cdot 90^{\circ} \\
=(4 n-1) \cdot 150^{\circ}+(n-1) \cdot 120^{\circ} \\
\leqslant(4 n-1) \cdot 150^{\circ} .
\end{array}
$$

Thus, $n \leqslant 1$, so $n=1$, meaning the given convex $4 n+2$-sided polygon is a convex hexagon. Therefore,
$$
\begin{array}{l}
\angle A_{4}+\angle A_{5}+\angle A_{6}=4 \cdot 180^{\circ}-3 \cdot 90^{\circ} \\
=450^{\circ} .
\end{array}
$$

Since $\angle A_{4}, \angle A_{5}, \angle A_{6} \leqslant 150^{\circ}$,
$$
\therefore \angle A_{4}=\angle A_{5}=\angle A_{6}=150^{\circ} \text {. }
$$

Thus, the degree measures of the internal angles are $\angle A_{1}=$
$$
\begin{array}{l}
\angle A_{2}=\angle A_{3}=90^{\circ}, \angle A_{4}=\angle A_{5}=\angle A_{6}= \\
150^{\circ} .
\end{array}
$$