Question

已知函数f(x)=2√3sinxcosx+2cosx^2-1

Answer 1

解f(x)=2√3sinxcosx+2cosx^2-1
=√3*2sinxcosx+（2cosx^2-1）
=√3sin2x+cos2x
=2（√圆拿3/搏腔磨2sin2x+1/2cos2x)
=2sin（2x+π/6)
T=2π/基斗2=π

Answer 2

答案如下乱拿兆：哗租敏颤f(x)=2√3sinxcosx+2cosx^2-1=√3*2sinxcosx+（2cosx^2-1=√3sin2x+cos2x
=2（√3/2sin2x+1/2cos2x)=2sin（2x+π/6) T=2π/2=π

Answer 3

f(x)=2√3sinxcosx+2cosx^2-1=√前烂轿3*2sinxcosx+（2cosx^2-1）慧肆=√3sin2x+cos2x=2（√3/2sin2x+1/2cos2x)=2sin（2x+π/6)
T=2π/历孝2=π