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(2014•贵阳模拟)已知函数f(x)=sin(2x+π3)-12(0≤x≤4π3)的零点为x1,x2,x3(x1<x2

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(2014•贵阳模拟)已知函数f(x)=sin(2x+
π
3
(2014•贵阳模拟)已知函数f(x)=sin(2x+π3)-12(0≤x≤4π3)的零点为x1,x2,x3(x1<x2
令函数f(x)=sin(2x+
π
3)-
1
2=0,求得sin(2x+
π
3)=
1
2,∴2x+
π
3=2kπ+
π
6,或 2x+
π
3=2kπ+

6,k∈z.
求得x=kπ-
π
12,或 x=kπ+
π
4.
再结合 x1<x2<x3,且x1、x2、x3∈[0,

3],可得x1 =
π
4,x2 =
11π
12,x3=

4,
∴cos(x1+2x2+x3)=cos
10π
3=cos

3=-cos
π
3=-
1
2,
故选:B.