Three discriminants of the equations are denoted as $\Delta_{1}, \Delta_{2}, \Delta_{3}$, respectively, then
$$\begin{array}{l} \Delta_{1}=\left(1-\cos ^{3} \alpha\right)^{2}-4 \cos \alpha, \\ \Delta_{2}=\left(1-\sin ^{3} \alpha\right)^{2}-4 \sin \alpha, \\ \Delta_{3}=\left(\sqrt{\frac{1+\cos \alpha}{1-\sin \alpha}}\right)^{2}-1 . \end{array}$$

Obviously, if $\cos \alpha > 0$, then $\Delta_{1} > 0$. Therefore, the equation $x^{2}-\left(1-\cos ^{3} \alpha\right) x+\cos \alpha=0$ has two distinct real roots.

If $\sin \alpha > 0$, then $\Delta_{2} > 0$. Therefore, the equation $x^{2}-$ $\left(1-\sin ^{3} \alpha\right) x+\sin \alpha=0$ has two distinct real roots.

If $\cos \alpha \geqslant 0$ and $\sin \alpha \geqslant 0$, by the given $\alpha \neq \frac{\pi}{2}+2 k \pi$ $(k \in \mathbf{Z}), 1-\sin \alpha \neq 0$, then $\sqrt{\frac{1+\cos \alpha}{1-\sin \alpha}}>1$. So,
$$\Delta_{3}=\left(\sqrt{\frac{1+\cos \alpha}{1-\sin \alpha}}\right)^{2}-1>0 \text {, }$$

In this case, the equation $x^{2}-\sqrt{\frac{1+\cos \alpha}{1-\sin \alpha}} x+\frac{1}{4}=0$ has two distinct real roots.

In summary, at least one of the three given equations has two distinct real roots.