1. Uso de reducción de potencia:
$\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$.
$$ \frac{1 + \cos 2x}{2} + \frac{1 + \cos 2y}{2} = \frac{1}{4} \implies 2 + \cos 2x + \cos 2y = \frac{1}{2} $$
$$ \cos 2x + \cos 2y = -\frac{3}{2} $$

2. Transformación a producto:
$$ 2 \cos(x+y) \cos(x-y) = -\frac{3}{2} $$
Sustituyendo $x+y = \frac{5\pi}{6}$:
$$ 2 \cos\left(\frac{5\pi}{6}\right) \cos(x-y) = -\frac{3}{2} \implies 2 \left(-\frac{\sqrt{3}}{2}\right) \cos(x-y) = -\frac{3}{2} $$
$$ -\sqrt{3} \cos(x-y) = -\frac{3}{2} \implies \cos(x-y) = \frac{\sqrt{3}}{2} $$

3. Hallando ángulos:
$x-y = \pm \frac{\pi}{6} + 2k\pi$. Sumando y restando con $x+y = \frac{5\pi}{6}$:
Si $x-y = \frac{\pi}{6} \implies 2x = \pi \implies x = \frac{\pi}{2}, y = \frac{\pi}{3}$.
Si $x-y = -\frac{\pi}{6} \implies 2x = \frac{4\pi}{6} \implies x = \frac{\pi}{3}, y = \frac{\pi}{2}$.

Resultado final:
$$ \boxed{(x, y) = \left(\frac{\pi}{2} + k\pi, \frac{\pi}{3} - k\pi\right), \quad \left(\frac{\pi}{3} + k\pi, \frac{\pi}{2} - k\pi\right)} $$