函数f(x)=cos(2x+3π）+sin²x ;求函数的最大值和最小正周期

f(x)=cos(2x+π/3)+sin²x=cos2xcos(π/3)-sin2xsin(π/3)+(1-cos2x)/2=cos2x/2-√3sin2x/2-cos2x/2+1/2=-(√3/2)sin2x+1/2最大值是:f(-π/4)=cos[2(-π/4)+π/3]+sin²(-π/4)=√3/2+1/2=(1+√3)/2≈1.366...