f(x)=2sinxcosx+2√3cos²X-1-√3=sin2x+√3(2cos²X-1)-1=sin2x+√3cos2x-1=2sin(2x+π/3)-1由2x+π/3=kπ,k∈Z得：x=kπ/2-π/6,k∈Z∴ f（x）的對(duì)稱中心為（kπ/2-π/6,-1）,k∈Z由2x+π/3=kπ+π/2,k∈Z...求（1）當(dāng)x屬于[0.π/2]時(shí)，求f(x)的最大值及此時(shí)x的值。（3）求f(x)在x屬于[-2/π，2/π]上的單調(diào)遞增區(qū)間。（4）把f（x）的圖像享有平移個(gè)單位后。所得圖像關(guān)于y軸對(duì)稱。求m的最小值。（5）當(dāng)x屬于[0,π]時(shí),求f（x）=0的x值。（6）說明怎樣由y=sinx的圖像得到函數(shù)f(x)的圖像。1∵x∈[0.π/2],2x+π/3∈[π/3,4π/3] ∴當(dāng) 2x+π/3=π/2,x=π/12,f(x)max=13.2kπ-π/2≤2x+π/3≤ 2kπ+π/2kπ-5π/12≤x+≤ kπ+π/12 k=0 時(shí)得f(x)在x屬于[-π/2，π/2]上 單調(diào)遞增區(qū)間 [-5π/12,π/12]4 f(x)=2sin(2x+π/3)-1對(duì)稱軸方程為x=kπ/2-π/12,k∈Z k=0時(shí)，x=-π/12，距y軸最近f(x)向左移π/12即可使圖像關(guān)于y軸對(duì)稱，最小平移量π/125. 2sin(2x+π/3)-1=0 ==>sin(2x+π/3)=1/2 ∵2x+π/3∈[π/3,7π/3]∴2x+π/3=5π/6 ,==> x=π/4 2x+π/3=13π/6 ==>x=11π/126.y=sinx---> (1)y=sin(x+π/3)--->（2） y=sin(2x+π/3)--->（3） y=sin(2x+π/3)-1(1)左移π/3 （2）縱不變，橫坐標(biāo)變?yōu)?/2倍 （3）下移1個(gè)單位