Ответ: Решениеx — y = π/2cosx — cosy = — √2x = π/2 + y cos(π/2 + y) — cosy = — √2- siny — cosy = — √2-(cosy + siny) = — √2- (√2cos(π/4 — y) = — √2cos(y — π/4) = 1y — π/4 = 2πk, k∈Zy = π/4 + 2πk, k∈Zx = π/2 + π/4 + 2πk, k∈Zx = 3π/4 + 2πk, k∈ZОтвет: x = 3π/4 + 2πk, k∈Z ; y = π/4 + 2πk, k∈Z
Ответ: { 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.