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...
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
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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)
=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)
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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
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
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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}
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}
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(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
=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
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2012-05-26
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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
=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
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