√(1+sinx)/cosx=1
√(1+sinx)=cosx ОДЗ: cosx≠0 x≠π/2+πn
1+sinx=cosx
sin²(x/2)+cos²(x/2)+2*sin(x/2)*cos(x/2) -cos²(x/2)+sin²(x/2)=0
2sin²(x/2)+2sin(x/2)*cos(x/2)=0
2*sin(x/2)*(sin(x/2)+cos(x/2))=0
sin(x/2)=0
x/2=πn
x₁=2πn
sin(x/2)+cos(x/2)=0
sin(x/2)=-cos(x/2) cos(x/2)≠-1 x/2≠-π/2+2πn x≠π+2πn
tg(x/2)=-1
x/2=-π/4+πn
x₂=-π/2+2πn x₂∉ по ОДЗ
Ответ: x₁=2πn.