Question

已知函数f(x)=sin（x+π/6)+sin(x-π/6)+cosx+a的最大值为1.

已知函数f(x)=sin（x+π/6)+sin(x-π/6)+cosx+a的最大值为1.
（1）求函数a的值。
（2）求使f(x)>=0成立的x的取值集合。

Answer 1

f(x)=sinxcosπ/6+cosxsinπ/6+2sinxcosπ/6-cosxsinπ/6+cosx+a
=2sinxcosπ/6+cosx+a
=√3sinx+cosx+a
=2sin(x+π/6)+a
最大值=2+a=1
a=-1

f(x)=2sin(x+π/6)-1>=0
sin(x+π/6)>=1/2=sinπ/6=sin5π/6
所以2kπ+π/6<=x+π/6<=2kπ+5π/6
2kπ<=x<=2kπ+2π/3
所以x∈{x|2kπ<=x<=2kπ+2π/3,k∈Z}

Answer 2

a=-1
x大于2kπ小于π|3+2kπ