函数f(x)=[2sin^2 (x+π/4)-1]- √3cos2x+1
=[-cos(2x+π/2)]- √3cos2x+1
=sin2x- √3cos2x+1
=2(1/2*sin2x-√3/2*cos2x)+1
=2sin(2x-π/3)+1
所以函数f(x)的最大值是3,最小值是-1.
最小正周期是π.
已知函数f(x)=2sin^2（π/4+x）-根号3cos2x
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