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