已知函数f(x)=sin(2x+π/3)-√3sin^2x+sinxcosx+√3/2
1个回答

sin(2x+π/3)-√3sin²x+sinxcosx+√3/2

=sin(2x+π/3) - 2sinx【(√3/2)sinx -(1/2)cosx】+(√3/2)

=sin(2x+π/3) - 2sinx【sin(π/3)sinx - cos(π/3)cosx】+(√3/2)

=sin(2x+π/3) + 2 sinx cos(x+π/3)+(√3/2)

=sin(2x+π/3) + 【sin(2x+π/3)+sin(π/3) 】+(√3/2)

=2sin(2x+π/3) +(√3/2)+(√3/2)

=2sin(2x+π/3) + √3

最小正周期为π

最小值为-2+√3,此时的 x =(k+7/12)π