1+sinx=sin^2(x/2)+cos^2(x/2)+2sin(x/2)cos(x/2)=(sin(x/2)+cos(x/2))^2;
1+cosx=1+2cos(x/2))^2-1=2*cos(x/2)^2;
所以,(1+sinx)/(1+cosx)=1/2((sin(x/2)+cos(x/2))^2/cos(x/2)^2)=1/2[(sin(x/2)+cos(x/2))/cos(x/2)]^2
=1/2(tan(x/2)+1)^2=1/2;
所以tan(x/2)=0,x/2=k*pi,x=2k*pi.k∈Z