1+cosx-ctgx/2 1+cosx=2cos^2(x/2) => 2cos^2(x/2)=ctgx/2 =>2cos^2(x/2)-cos(x/2)/sin(x/2)=0 cos(x/2)*(2cos(x/2)-1/sin(x/2)=0
cos(x/2)=0 2cos(x/2)-1/sin(x/2)=0 (2cos(x/2)*sin(x/2)-1)/sin(x/2)=0 sin(x/2)не=0 x/2не=pi*n x не=2pi*n n(- Z
cos(x/2)=0 x/2=pi/2+pi*n x=pi+2pi*n ,
2cos(x/2)*sin(x/2)-1=0 sinx-1=0 sinx=1 x=pi/2+2pi*n n принадлежит Z