已知函数f(x)=2√3sinxcosx+2cos²x-1

问题描述:

已知函数f(x)=2√3sinxcosx+2cos²x-1
(1)求函数f(x)的对称轴及在闭区间(0,π/2)上的最值
(2)若f(x0)=6/5,x0∈闭区间(π/4,π/2)求cos2x0的值

f(x)=2√3sinxcosx+2cos²x-1
=√3sin2x+cos2x
=2(√3/2sin2x+1/2cos2x)
=2(sin2xcosπ/6+cos2xsinπ/6)
=2sin(2x+π/6)
(1)
对称轴:2x+π/6=2kπ+π/2
2x=2kπ+π/3
x=kπ+π/6;k∈Z
闭区间【0,π/2】
当x=π/2时;函数有最小值=-2sin(π/6)=-1
当x=π/6时,函数有最大值=2sin(π/2)=-2
(2)2sin(2x0+π/6)=6/5
sin(2x0+π/6)=3/5
sin2x0cosπ/6+cos2x0sinπ/6=3/5
√3sin2x0+cos2x0=6/5 (1)
x0∈[π/4,π/2]
2x0∈[π/2,π]
cos2x0≤0
2x0+π/6∈[2π/3,7π/6]
因为:sin(2x0+π/6)=3/5>0
所以:2x0+π/6∈[2π/3,π)
所以:cos(2x0+π/6)