Question

сos2x=cosx+sinx
Решите уравнение

Answer 1

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.