4-14 Try to find all real roots of $x^{2}-2 x \sin \frac{\pi x}{2}+1=0$.


[Solution] Since $x$ is a real number, the discriminant of the given quadratic equation should be greater than or equal to zero. $\left(2 \sin \frac{\pi x}{2}\right)^{2}-4 \geqslant 0$, which means $\left(\sin \frac{\pi x}{2}\right)^{2} \geqslant 1$.

However, we also have $\left(\sin \frac{\pi x}{2}\right)^{2} \leqslant 1$,
so $\left(\sin \frac{\pi x}{2}\right)^{2}=1$,
thus $\sin \frac{\pi x}{2}= \pm 1$.
Therefore, $x= \pm 1, \pm 3, \pm 5, \cdots$
Substituting into the original equation, only $x= \pm 1$ satisfies, so the real roots sought are $x= \pm 1$.