Verified answer

y'=cosx*cos^2(x/2)-sinx*2cos(x/2)*sin(x/2)1/2=cosxcos^2(x/2)-sin^2x/2

y'=0

1/2cosx*(1+cosx)-(1-cos^2x)=0

1/2cosx+1/2cos^2x-1+cos^2x=0

3/2cos^2x+1/2cosx-1=0

cosx=t

3t^2+t-2=0

t1=-1  x=П

t2=2/3 х=arccos2/3

y(П)=0 - минимум

cosx=2/3  cos^2x/2=1/2(1+2/3)=5/6

sinx=sqrt(1-4/9)=sqrt(5)/3

y=(5/18)*sqrt(5)- максимум