сos2x=cosx+sinx
Решите уравнение

cos2x = cosx + sinxcos²x - sin²x = cosx + sinx(cosx - sinx)(cosx + sinx) - (cosx + sinx) = 0(cosx - sinx - 1)(cosx + sinx) = 0 cosx - sinx - 1 = 0cosx - sinx = 1      |:√2cosx/√2 - sin/√2 = √2/2cosx·cos(π/4) - sinx·sin(π/4) = √2/2cos(x + π/4) = √2/2x + π/4 = ±π/4 + 2πk, k ∈ Zx = ±π/4 - π/4 + 2πk, k ∈ Zcosx - sinx = 0sinx = cosxtgx = 1x =π/4 + πn, n ∈ ZОтвет: x = ±π/4 - π/4 + 2πk, k ∈ Z; π/4 + πn, n ∈ Z.