1. Identidades de reducción de potencia:
$\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$. Sustituimos en la primera ecuación:
$$ \frac{1 - \cos 2x}{2} + \frac{1 - \cos 2y}{2} = \frac{3}{4} \implies 2 - (\cos 2x + \cos 2y) = \frac{3}{2} $$
$$ \cos 2x + \cos 2y = \frac{1}{2} $$

2. Transformación a producto:
$$ 2 \cos(x+y) \cos(x-y) = \frac{1}{2} $$
Sustituyendo $x+y = \frac{\pi}{3}$:
$$ 2 \cos\left(\frac{\pi}{3}\right) \cos(x-y) = \frac{1}{2} \implies 2 \left(\frac{1}{2}\right) \cos(x-y) = \frac{1}{2} $$
$$ \cos(x-y) = \frac{1}{2} \implies x-y = \pm \frac{\pi}{3} + 2k\pi $$

3. Resolución para $k=0$:
Si $x-y = \frac{\pi}{3}$ y $x+y = \frac{\pi}{3} \implies x = \frac{\pi}{3}, y = 0$.
Si $x-y = -\frac{\pi}{3}$ y $x+y = \frac{\pi}{3} \implies x = 0, y = \frac{\pi}{3}$.

Resultado final:
$$ \boxed{x_1 = \frac{\pi}{6} + k\pi \pm \frac{\pi}{6}, \quad y_1 = \frac{\pi}{6} + k\pi \mp \frac{\pi}{6}} $$