解题思路:由f(x)=sin(2x+[π/3])-[1/2]=0⇒sin(2x+[π/3])=[1/2],依题意,可求得x1=[π/4],x2=[11π/12],x3=[5π/4],从而可求得cos(x1+2x2+x3)的值.
由f(x)=sin(2x+[π/3])-[1/2]=0得:sin(2x+[π/3])=[1/2],
∵0≤x≤[4π/3],
∴[π/3]≤2x+[π/3]≤3π,
∴2x+[π/3]=[5π/6]或2x+[π/3]=[13π/6]或2x+[π/3]=[17π/6],
解得x1=[π/4],x2=[11π/12],x3=[5π/4],
∴x1+2x2+x3=[10π/3],
∴cos(x1+2x2+x3)=cos[10π/3]=cos(4π-[2π/3])=cos(-[2π/3])=cos[2π/3]=-[1/2].
故答案为:-[1/2].
点评:
本题考点: 函数y=Asin(ωx+φ)的图象变换.
考点点评: 本题考查正弦函数的图象与性质,求得x1=[π/4],x2=[11π/12],x3=[5π/4]是关键,考查函数零点的应用,考查运算求解能力,属于中档题.