已知函数f(x)=sin(2x+[π/3])-[1/2](0≤x≤[4π/3])的零点为x1、x2、x3(x1<x2<x

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  • 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.