Помогите решить cos2x=√2 (cos x-sin x)

cos2x=√ √2 (cos x-sin x)

cos2x - √2 (cos x-sin x) = 0

Сos ²x - Sin²x - √2 (cos x-sin x) = 0

(Cosx - Sin x) (Cosx + Sin x) - √2 (cos x-sin x) = 0

(Cos x - Sin x) (Cos x + Sinx - √2) = 0

Cos x - Sin x = 0 | : Сos x ≠0 или Сos x + Sin x - √2 = 0

1 - tg x = 0 (1-tg²x/2) / (1 + tg²x/2) + 2tgx/2 / (1 + tg²x/2) = √2

tgx = 1 tgx/2 = 0 или tgx/2 = 1

x = π/4 + πk, k ∈Z x/2 = πn, n ∈Z x/2 = π/4 + πm, m ∈Z

х = 2πn, n ∈Z x = π/2 + 2πm, m∈Z