求sin2x+2根号3sin²x的最小正周期
问题描述：
求sin2x+2根号3sin²x的最小正周期
答
原式=
=sin2x+√3(1-cos2x)
=sin2x-√3cos2x+√3
=2sin(2x+π/6)+√3
最小正周期为t=2π/2=π
答
sin(2x)＋2√3sin²x
=sin(2x)＋√3×[1－cos(2x)]
=sin(2x)－√3cos(2x)＋√3
=2sin(2x－π/3)＋√3
最小正周期是2π/2=π