РЕШИТЕ СИСТЕМУ УРАВНЕНИЙ: [latex] \left \{ {{cosx+cosy= \sqrt{3} } \atop {x+y= \pi/3}} \right. [/latex]

[latex] \left\{\begin{array}{l} \cos{x} + \cos{y} = \sqrt{3} \ , \\\\ x + y = \frac{ \pi }{3} \ ; \end{array}\right [/latex] [latex] \left\{\begin{array}{l} 2 \cos{ \frac{ x + y }{2} } \cos{ \frac{ x - y }{2} } = \sqrt{3} \ , \\\\ y = \frac{ \pi }{3} - x \ ; \end{array}\right [/latex] [latex] \left\{\begin{array}{l} \left|\begin{array}{l} 2 \cos{ \frac{ \pi/3 }{2} } \cos{ \frac{ x - [ \pi/3 - x ] }{2} } = \sqrt{3} \ , \\\\ 2 \cos{ \frac{ \pi }{6} } \cos{ \frac{ 2x - \pi/3 }{2} } = \sqrt{3} \ , \\\\ 2 \cdot \frac{ \sqrt{3} }{2} \cos{ ( x - \frac{ \pi }{6} ) } = \sqrt{3} \ , \\\\ \sqrt{3} \cos{ ( x - \frac{ \pi }{6} ) } = \sqrt{3} \ , \\\\ \cos{ ( x - \frac{ \pi }{6} ) } = 1 \ , \\\\ ( x - \frac{ \pi }{6} ) = 2 \pi n \ , \ n \in Z \ ; \end{array}\right \\\\ y = \frac{ \pi }{3} - x \ ; \end{array}\right [/latex] О т в е т :    [latex] \left\{\begin{array}{l} n \in Z \ , \\ x = \frac{ \pi }{6} + 2 \pi n \ , \\ y = \frac{ \pi }{6} - 2 \pi n \ ; \end{array}\right [/latex]

{cosx +cosy =√3 ; x+y =π/3 .⇔{2cos(x+y)/2*cos(x-y)/2=√3 ; x+y =π/3.⇔ {2cosπ/6*cos(x-y)/2=√3 ; x+y =π/3.⇔{cos(x-y)/2=1 ; x+y =π/3.⇒  {x+y =π/3 ; x-y =4πn ,n∈Z .  || + и - ||   { x = π/6 +2πn ;y =π/6 -2πn ,n∈Z.