найдите все решения уравнения (sin x + cos x)^2=1 sin x

Sinx = cosx
2sin(x/2)cos(x/2) = cos^2(x/2) - sin^2(x/2) :cos^2(x/2)
tg^2(x/2) +2tg(x/2) - 1 = 0
tg(x/2) = t
t^2 +2t - 1 = 0
D = 4 + 4 = 8
t = (-2 +- 8^0,5)/2 = -1 +- 2^0,5
tg(x/2) = -1 + 2^0,5 tg(x/2) = -1 - 2^0,5
x/2 = arctg(2^0,5 - 1) +Пk x/2 = arctg(-1 - 2^0,5) +Пk
x = 2arctg(2^0,5 - 1) + 2Пk x = -2arctg(1 + 2^0,5) + 2Пk

[-2п;0]
x = 2arctg(2^0,5 - 1) - 2П
x = -2arctg(1 + 2^0,5)
(Немного не уверена)