f(x)=sin2x-2√3sin^2x+√3+1,则函数f(x)的最小正周期是?麻烦帮我化简并解答,要准确步骤,
问题描述：
f(x)=sin2x-2√3sin^2x+√3+1,则函数f(x)的最小正周期是?麻烦帮我化简并解答,要准确步骤,
答
√3sin请问这是根号？一直括打哪？
答
f(x)=sin2x-2√3sin^2x+√3+1=sin2x-2√3*（1-cos2x）/2+√3+1=sin2x-√3+√3cos2x+√3+1=sin2x+√3cos2x+1=2sin(2x+π/3)+1 所以最小正周期为π
答
f(x)=sin2x-2√3sin^2x+√3+1
=sin2x-√3(1-cos2x)+√3+1
=sin2x+√3cos2x+1
=2sin(2x+60°）+1
函数f(x)的最小正周期是PAI
答
f(x)=sin2x-2√3sin²x+√3+1
=sin2x-2√3(1-cos2x)/2+√3+1
=sin2x+√3cos2x+1
=2(1/2sin2x+√3/2cos2x)+1
=√2sin(2x+π/3)+1
最小正周期为2π/2=π
答
恩.
因为2sin^2x=1-cos2x
所以原式=sin2x-√3(1-cos2x)+√3+1
所以原式=sin2x+√3cos2x+1
所以原式=2sin(2x+π∕3）+1
周期为π