428. Solve the equation:

$$
x^{2}-2 x \sin y+1=0
$$
$\triangle$ Let's represent 1 on the left side of the equation as $\sin ^{2} y+\cos ^{2} y$. Then

$$
\left(x^{2}-2 x \sin y+\sin ^{2} y\right)+\cos ^{2} y=0, \quad(x-\sin y)^{2}+\cos ^{2} y=0
$$

Therefore,

$$
x-\sin y=0, \quad \cos y=0
$$

If $\cos y=0$, then

$$
\sin y= \pm 1, \quad x=\sin y= \pm 1
$$

Let's consider two cases.

1) Suppose $\sin y=1$. Then

$$
y=\frac{\pi}{2}+2 \pi k(k \in Z), x=1
$$

2) Suppose $\sin y=-1$. Then

$$
y=-\frac{\pi}{2}+2 \pi n(n \in Z), x=-1
$$

Answer: $(1 ; \pi / 2+2 \pi k)(k \in Z) ;(-1 ;-\pi / 2+2 \pi n)(\mathrm{n} \in Z)$.