f(x)=2√3sinxcosx+2cos²x-1
=√3sin2x+cos2x
=2(√3/2sin2x+1/2cos2x)
=2(sin2xcosπ/6+cos2xsinπ/6）
=2sin(2x+π/6)
(1)
對(duì)稱軸：2x+π/6=2kπ+π/2
2x=2kπ+π/3
x=kπ+π/6；k∈Z
閉區(qū)間【0,π/2】
當(dāng)x=π/2時(shí)；函數(shù)有最小值=-2sin(π/6)=-1
當(dāng)x=π/6時(shí),函數(shù)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?yàn)椋簊in(2x0+π/6)=3/5>0
所以：2x0+π/6∈[2π/3,π)
所以：cos(2x0+π/6)