Question

已知函数f(x)=cos(x+x/6)-sin(x-2π/3)+sinx+a的最大值为1。求常数a

(1)求常数a的值
(2)求使f(x)≥0成立的x的取值范围

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Answer 1

f(x)=cos(x+π/6)-sin(x-2π/3)+sinx+a
=√3/2cosx-1/2sinx+1/2sinx+√3/2cosx+sinx+a
=√3cosx+sinx+a
=2(√3/2cosx+1/2sinx)+a
=2sin(x+π/3)+a
2+a=1
a=-1
2)f(x)=2sin(x+π/3)-1≥0

sin(x+π/3)≥1/2
2kπ+5π/6≥x+π/3≥2kπ+π/6
2kπ+π/2≥x+π/3≥2kπ-π/6