Сos2x=cosx+sinx

Решите уравнение

Cos2x = cosx + sinx

cos²x - sin²x = cosx + sinx

(cosx - sinx) (cosx + sinx) - (cosx + sinx) = 0

(cosx - sinx - 1) (cosx + sinx) = 0

cosx - sinx - 1 = 0

cosx - sinx = 1 |:√2

cosx/√2 - sin/√2 = √2/2

cosx·cos (π/4) - sinx·sin (π/4) = √2/2

cos (x + π/4) = √2/2

x + π/4 = ±π/4 + 2πk, k ∈ Z

x = ±π/4 - π/4 + 2πk, k ∈ Z

cosx - sinx = 0

sinx = cosx

tgx = 1

x = π/4 + πn, n ∈ Z

Ответ: x = ±π/4 - π/4 + 2πk, k ∈ Z; π/4 + πn, n ∈ Z.