=-√3cos2x- cos(2x+π/2)
=-√3cos2x+sin2x
=2sin(2x-π/3)
最小值為-2,最大值為2
最小正周期為T=π
2)h(x)=f(x+t)=2sin(2x+2t-π/3)關(guān)于點(diǎn)（-π/6,0)對稱
則2sin(2×(-π/6)+2t-π/3)=0
故2×(-π/6)+2t-π/3=kπ,t屬于（0,π）,
所以t= kπ/2+π/3,k∈Z.
t=π/3或5π/6.設(shè)p:x屬于[π/4，π/2]，q：f（x）-m的絕對值<3，若p是q的充分條件，求實(shí)數(shù)m的取值范圍。謝謝。。當(dāng)x屬于[π/4,π/2]時，f(x) =2sin(2x-π/3)2x-π/3屬于[π/6,2π/3]sin(2x-π/3)屬于[1/2,1]所以f(x) 屬于[1,2]若p是q的充分條件，則有If(x)-mI<3-3