请帮忙解答一下 已知函数f(x)=2cosx*sin(x+π/3)-√3sin平方x+sinx*cosx 1 求函数f(x)的最小正周期2 求f(x)的最小值及取的最小值相应的x值？请用高一的知识回答，请问平方怎么打

解 答案是π
f(x)=2cosx（sinx/2+√3cosx/2)-√3sin平方x+sinxcosx
=sinxcosx+√3cos平方x-√3sin平方x+sinxcosx
=2sinxcos+√3(cos平方x-sin平方x)
=sin2x+√3cos2x
=2sin（2x+π/3）
T=2π/2=π

先化简函数：=2cosx*sin(x+π/3)+2sinx *cos(x+π/3)
=2sin(2x+π/3)
所以最小正周期是π。

f(x)=2cosxsin(x+π/3)-√3sin^2x+sinxcosx
=2cosx[(1/2)sinx+(√3/2)cosx]+(√3/2)(1-2sin^2x)-(√3/2)+(1/2)sin2x
=sinxcosx+√3cos^2x+(√3/2)cos2x-(√3/2)+(1/2)sin2x
=(1/2)sin2x+(√3/2)(2cos^2x-1)+(√3/2)+(√3/2)cos2x-(√3/2)+(1/2)sin2x
=sin2x+√3cos2x
=2*[(1/2)sin2x+(√3/2)cos2x]
=2sin(2x+π/3)
最小正周期：2π/ω=2π/2=π
f(x)的最小值：-2
当2x+π/3=2kπ-(π/2)时，取得最小值
x=kπ-(5π/12) ，k∈Z

f(x)=2cosx*sin(x+π/3)-√3﹙sinx﹚^2+sinx*cosx
=2cosx*﹙sinxcosπ/3+cosxsinπ/3﹚-√3﹙sinx﹚^2+sinx*cosx
=cosxsinx+√3﹙cosx﹚^2-√3﹙sinx﹚^2+sinx*cosx
=sin2x+√3cos2x
=2[1/2sin2x+√3/2cos2x]
=2[sin﹙2x+π/3﹚]
函数f(x)的最小正周期T=2π/ω=π

1. f(x)=2cosx*sin(x+π/3)-√3﹙sinx﹚^2+sinx*cosx =2cosx*﹙sinxcosπ/3+cosxsinπ/3﹚-√3﹙sinx﹚^2+sinx*cosx=cosxsinx+√3﹙cosx﹚^2-√3﹙sinx﹚^2+sinx*cosx=sin2x+√3cos2x=2sin﹙2x+π/3﹚∴函数f(x)的...