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

 { 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.

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