428 а плиииииииииииииииииз

$\tt\displaystyle \left \{ {{x-y=\frac{\pi}{6}} \atop {\cos x\sin y=\frac{1}{4}}} \right. \left \{ {{x-y=\frac{\pi}{6}} \atop {\frac{1}{2}(\sin (x+y)+sin(x-y))=\frac{1}{4}|\cdot 2}} \right. \left \{ {{x-y=\frac{\pi}{6}} \atop {\sin (x+y)+\sin \frac{\pi}{6}=\frac{1}{2}}} \right. \\\\\\\left \{ {{x-y=\frac{\pi}{6}} \atop {\sin (x+y)=0}} \right. \left \{ {{x-y=\frac{\pi}{6}} \atop {x+y=\pi n, n\in Z}} \right. |+\left \{ {{2x=\frac{\pi}{6}+\pi n,n\in Z} \atop {\frac{\pi}{6}+y+y=\pi n,n\in Z}} \right.$

$\tt\displaystyle \left \{ {{x=\frac{\pi}{12}+\frac{\pi n}{2}, n\in Z} \atop {y=-\frac{\pi}{12}+\frac{\pi n}{2}, n\in Z}} \right.$

Ответ: $(\frac{\pi}{12}+\frac{\pi n}{2};-\frac{\pi}{12}+\frac{\pi n}{2});n\in Z$