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）求常数a的值。 （2）求使f(x)大于等于零x值
按您想法的讲讲

Answer 1

f(x)=sin(x+π/6)+sin(x-6π)+cos(x)+a
=sin(x+π/6)+sin(x)cos(π/6)-cos(x)sin(π/6)+cos(x)+a
=sin(x+π/6)+sin(x)cos(π/6)+cos(x)sin(π/6)+a
=2sin(x+π/6)+a
当sin(x+π/6)=1时，f(x)的最大值为2+a=1，可得a=-1
得f(x)=2sin(x+π/6)-1，由f(x)≥0得sin(x+π/6)≥1/2
可得2kπ+π/6≤x+π/6≤2kπ+5π/6
即2kπ≤x≤2kπ+2π/3
即使f(x)≥0成立的x的取值集合为[2kπ，2kπ+2π/3]，(k∈Z)

Answer 2

解1：
f(x)=sin(x+π/6)+sin(x-π/6)+cosx+a
f(x)=2sinxcos(π/6)+cosx+a
f(x)=√3sinx+cosx+a
f(x)=2[(√3/2)sinx+(1/2)cosx]+a
f(x)=2[cos(π/6)sinx+sin(π/6)cosx]+a
f(x)=2sin(x+π/6)+a
可见：f(x)≤2+a
因为：f(x)≤1
所以：2+a=1
解得：a=-1
解2：
由上解，有：f(x)=2sin(x+π/6)-1
因为：f(x)≥0
所以：2sin(x+π/6)-1≥0
即：sin(x+π/6)≥1/2
有：2kπ+5π/6≥x+π/6≥2kπ+π/6，k∈Z
解得：2kπ+2π/3≥x≥2kπ，k∈Z

Answer 3

y=2sinxcosπ/6+cosx+a=根号3sinx+cosx+a=2sin(x+π/6)+a
2sin(x+π/6)的最大值为2因为ymax=2+a=1
所以a=-1
所以y>=0->sin(x+π/6)>=1/2
所以2kπ+π/6<=x+π/6<=2kπ+5π/6
所以2kπ<=x<=2kπ+2π/3
集合为{x|2kπ<=x<=2kπ+2π/3,k属于Z}

Answer 4

(1) f(x)=sinxcosπ/6+cosxsinπ/6+sinxcosπ/6-cosxsinπ/6+cosx+a
=2sinxcosπ/6+cosx+a
=√3sinx+cosx+a
=2sin(x+π/4)+a
最大值=2+a=1 a=-1
(2)
2sin(x+π/4)-1>=0
sin(x+π/4)>=1/2
2kπ+π/6<=x+π/4<=2kπ+5π/6
2kπ-π/12<=x<=2kπ+7π/12

Answer 5

f(x)=sinxcosπ/6+cosxsinπ/6+sinxcosπ/6-cosxsinπ/6+cosx+a
=2sin(x+π/4)+a
最大值2+a=1 a=-1

2sin(x+π/4)-1>=0
sin(x+π/4)>=1/2
2kπ-π/12<=x<=2kπ+7π/12