Answer: the number of self-intersections is $n$, where $n$ is the largest natural number for which $n \boldsymbol{\alpha}<180^{\circ}$. Consider points $S$ and $A$ on a plane, corresponding to the vertex of the cone and a chosen point on the plane when the cone is unfolded onto this plane. Construct a perpendicular from point $A$ to the line $S A$ and take points $B_{1}, \ldots, B_{\mathrm{n}}$ on this perpendicular on one side of the line $S A$ such that $\angle A S B_{\mathrm{k}}=k \frac{\alpha}{2}$. Similarly, take points $C_{1}, \ldots, C_{\mathrm{n}}$ on the other side of the line $S A$ such that $\angle A S C_{\mathrm{k}}=k \frac{\alpha}{2}$. On the cone, point $B_{\mathrm{k}}$ coincides with point $C_{\mathrm{k}}$; there are no other points of self-intersection of the drawn line.
On an infinite cone, the angle of which is equal to $\alpha$, a point is taken. From this point, a line is drawn in both directions such that after unfolding, it turns into straight segments. Determine the number of its self-intersections.