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