Помогите решить тригонометрическое уравнение) sinx+cosx=√2

sinx + cosx = √2 Возведем в квадрат: sin²x + 2sinxcosx + cos²x = 2 1 + 2sinxcosx = 2 sin2x = 1 2x = π/2 + 2πn, n ∈ Z x = π/4 + 2πn, n ∈ Z.

Делим обе части на √2: 1/√2 · sinx + 1/√2 · cosx = 1 1/√2 = sin(π/4) = cos(π/4) sinx·cos(π/4) + cosx·sin(π/4) = 1 sin (x + π/4) = 1 x + π/4 = π/2 + 2πn x = π/2 - π/4 + 2πn x = π/4 + 2πn