Решить систему: x-y=Pi/2 cosx-cosy = - корень из 2

Решение x - y = π/2 cosx - cosy = - √2 x = π/2 + y  cos(π/2 + y) - cosy = - √2 - siny - cosy = - √2 -(cosy + siny) = - √2 - (√2cos(π/4 - y) = - √2 cos(y - π/4) = 1 y - π/4 = 2πk, k∈Z y = π/4 + 2πk, k∈Z x = π/2 + π/4 + 2πk, k∈Z x = 3π/4 + 2πk, k∈Z Ответ: x = 3π/4 + 2πk, k∈Z ; y = π/4 + 2πk, k∈Z  

 { x - y =π/2 ; cosx - cosy =√2  ⇔   { x - y =π/2 ; - 2sin(x-y)/2*sin(x+y)/2  =√2 . { x - y =π/2 ; - 2sinπ/4*sin(x+y)/2  =√2 .  - 2sinπ/4*sin(x+y)/2  =√2 ; -2*(1/√2)*sin(x+y)/2 =√2 ; sin(x+y) = -1; x+y = π+2π*k  , k∈ Z . {x+y = π+2π*k  , k∈ Z ;   x-y =π/2 ⇔ {2x =π+2π*k +π/2 ; 2y = π+2π*k -π/2.  {x =3/4π+ π*k    ;  y = π/4+ π*k  , k ∈Z. ответ :   x =3/4π+ π*k , k ∈Z  ,   y = π/4+ π*k  , k ∈Z.