6.8. (CSSR, 56). Find all pairs of numbers \( x, y \in (0, \pi / 2) \) that satisfy the system

\[
\left\{\begin{array}{l}
\frac{\cos x}{\cos y}=2 \cos ^{2} y \\
\frac{\sin x}{\sin y}=2 \sin ^{2} y
\end{array}\right.
\]
6.8. By adding the equations of the original system, we get for any solution of the system

$$
\frac{\cos x \sin y + \sin x \cos y}{\cos y \sin y} = 2
$$

i.e.,

$$
\frac{\sin (x+y)}{\sin 2 y} = 1
$$

Therefore,

$$
\sin \left(\frac{x-y}{2}\right) \cos \left(\frac{x+3 y}{2}\right) = 0
$$

Since \( x, y \in (0 ; \pi / 2) \), either \( x = y \) or \( x + 3 y = \pi \). Further, by considering the difference of the two original equations, we have

$$
\frac{\sin (y-x)}{\sin 2 y} = \cos 2 y, \text{ i.e., } \sin (y-x) = \frac{1}{2} \sin 4 y.
$$

If \( x = y \), then \( \sin 4 y = 0 \), from which \( y = \pi / 4 \) (since \( y \in (0 ; \pi / 2) \)). Hence, \( x = y = \pi / 4 \). If \( x = \pi - 3 y \), then \( \sin (4 y - \pi) = -\frac{1}{2} \sin 4 y \), i.e., \( -\sin 4 y = \frac{1}{2} \sin 4 y \), so \( \sin 4 y = 0 \). Therefore, we again get \( y = \pi / 4, x = \pi - 3 \pi / 4 = \pi / 4 \). Thus, we have \( x = y = \pi / 4 \). By verification, we confirm that the values \( x = y = \pi / 4 \) indeed satisfy the original system.