Ответ:(2π/3+2πn; 2πn) (2πk; -2π/3+2πk) n и k принадлежат Z
Пошаговое объяснение:
{x-y=2π/3
{cosx+cosy=1/2
{x=2π/3+y
cos(2π/3+y)+cosy=1/2
cos2π/3·cosy-sin2π/3·siny+cosy=1/2
-1/2cosy-√3/2siny+cosy=1/2
1/2·cosy-√3/2·siny=1/2
sinπ/6·cosy-cosπ/6·siny=1/2
sin(π/6-y)=1/2
sin(-(y-π/6))=1/2
-sin(y-π/6)=1/2
sin(y-π/6)=-1/2
y-π/6=-π/6+2πn y=2πn x=2π/3+2πn n принадлежит Z
y-π/6=-5π/6+2πk y=-2π/3+2πk x=2π/3-2π/3+2πk=2πk k принадлежит Z