求函数f(x)=1/2sin^2x+cos^2x+根号3/4sin2x的最大值

f(x)=1/2[(1-cos2x)/2]+[(1+cos2x)/2]+根号3/4sin2x=3/4+cos2x/4+根号3/4sin2x=3/4+1/2sin(π/6+2x)
所以最大值为;3/4+1/2=5/4

f(x)=1/2sin^2x+cos^2x+√3/4sin2x
=1/2*1/2(1-cos2x)+1/2(1+cos2x)+√3/4sin2x
=3/4+1/4cos2x+√3/4sin2x
=3/4+1/2sin(2x+π/6)
最大值是sin(2x+π/6)=1时取得为5/4

f(x)=sin^2x/2+cos^2x+√3/4sin2x
=1/2+cos^2x/2+√3/4sin2x
=1/2+(cos2x+1)/4+√3/4sin2x
=1/2+1/4*cos2x+1/4+√3/4sin2x
=1/4*cos2x+√3/4sin2x+3/4
=1/2*(1/2*cos2x+√3/2sin2x)+3/4
=1/2*(sinπ/6*cos2x+cosπ/6sin2x)+3/4
=1/2*sin(π/6+2x)+3/4
-1-1/21/4所以函数f(x)=sin^2x/2+cos^2x+√3/4sin2x的最大值为：5/4