1、f(x)=4sin(π/2+x/2)cos(x/2+π/6)
=4cos(x/2)[√3/2*cos(x/2)-1/2*sin(x/2)]
=2√3*cos²(x/2)-2sin(x/2)cos(x/2)
=√3*[2cos²(x/2)-1]-sinx+√3
=√3cosx-sinx+√3
=2(√3/2*cosx-1/2*sinx)+√3
=2cos(x+π/6)+√3
函數(shù)f(x)的周期為：T=2π/1=2π
令cos(x+π/6)=0
即x+π/6=kπ,k為整數(shù)
則x=kπ-π/6,k為整數(shù)
函數(shù)圖像對(duì)稱中心的坐標(biāo)為：(kπ-π/6,√3),k為整數(shù)

2、第二題請(qǐng)把題目敘述清楚!題目好像有問題