Помогите решить тригонометрическое уравнение) sinx+cosx=√2

sinx + cosx = √2Возведем в квадрат:sin²x + 2sinxcosx + cos²x = 21 + 2sinxcosx = 2sin2x = 12x = π/2 + 2πn, n ∈ Zx = π/4 + 2πn, n ∈ Z.

Делим обе части на √2:1/√2 · sinx + 1/√2 · cosx = 11/√2 = sin(π/4) = cos(π/4)sinx·cos(π/4) + cosx·sin(π/4) = 1sin (x + π/4) = 1x + π/4 = π/2 + 2πnx = π/2 - π/4 + 2πnx = π/4 + 2πn