x + y =2π
Sinx + Cosy = 1
решаем подстановкой: у = 2π - х
Sinx + Cos(2π -x) = 1
Sinx + Cosx = 1
Есть формулы: Sinx = 2tg(x/2) / (1 + tg²(x/2) )
Сosx = (1 -tg²(x/2) )/ (1 + tg²(x/2) )
Применим их
2tg(x/2) / (1 + tg²(x/2) ) + (1 -tg²(x/2) )/ (1 + tg²(x/2) ) = 1
2tg(x/2) + (1 -tg²(x/2) ) - (1 + tg²(x/2) ) = 0
2tg(x/2) + 1 -tg²(x/2) - 1 - tg²(x/2) = 0
-2 tg²(x/2) + 2tg(x/2) -2 = 0
tg(x/2) = t
t² + t = 0
t ₁= 0 t ₂= -1
tg(x/2) = 0 tg(x/2) = -1
x/2 = πk , k ∈ Z x/2 = -π/4 + πn , n ∈Z
x = 2πk , k ∈ Z x = -π/2 + 2πn , n ∈Z