求助一道高中数学题！

f(x)=cos(2x+π/3) +(sin^2)x=cos2x*1/2-sin2x*根号3/2+(1-cos2x)/2=-根号3/2*sin2x+1/2
所以,f(x)的最大值是-根号3/2*(-1)+1/2=(根号3+1)/2
(2)f(A/2)=-根号3/2sinA+1/2=-1/4.
sinA=根号3/2
即角A=120度或60度.
a^2=b^2+c^2-2bc*cosA
7=1+c^2-2c*(1/2)或者说7=1+c^2+2c*1/2
c^2-c-6=0或者说c^2+c-6=0
(c-3)(c+2)=0,或者说(c+3)(c-2)=0
c=3或者说c=2
S=1/2bcsinA=1/2*1*3*根号3/2=3根号3/4或者说S=1/2*1*2*根号3/2=根号3/2

f(x)=cos(2x+π/3) +(sin^2)x
=cos(2x)cos(π/3) - sin(2x)sin(π/3)+(sin^2)x
=1/2*(1 - 2(sin^2)x)-√3/2*sin(2x)+(sin^2)x
=1/2 - √3/2*sin(2x)
当sin(2x)=-1时，即x=π/4时，f(x)max=1/2 + √3/2；
(2)f(A/2)=1/2 -(√3/2)sinA=-4,
sinA=3√3>1;
题目给出的是错的

楼上正解