已知函数f(x)=sin(x+6分之派)+sin(x-6分之派)+cosx+a的最大值为1.1,求常数a的值2,求使f(x)大于等于0成立的x的取值集合.

Question

已知函数f(x)=sin(x+6分之派)+sin(x-6分之派)+cosx+a的最大值为1.1,求常数a的值 2,求使f(x)大于等于0成立的x的取值集合.

Answer 1

1
f(x)=sin(x+6分之派)+sin(x-6分之派)+cosx+a
=sinxcosπ/6+cosxsinπ/6+sinxcosπ/6-cosxsinπ/6+cosx+a
=2sinxcosπ/6+cosx+a
=√3sinx+cosx+a
=2sin(x+π/6)+a
∵f(x)最大值为1 ∴2+a=1,∴a=-1
2
f(x)=0,即2sin(x+π/6)-1=0
sin(x+π/6)=1/2
∴x+π/6=2kπ+π/6,或x+π/6=2kπ+5π/6,k∈Z
∴x=2kπ,或x=2kπ+2π/3,k∈Z
∴f(x)=0时,x集合为：
{x| x=2kπ,或x=2kπ+2π/3,k∈Z}