设函数fx=cos(2x-3分之4派)+2cos^2x (1)求fx最大值时x的集合(2

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设函数fx=cos(2x-3分之4派)+2cos^2x (1)求fx最大值时x的集合(2

设函数fx=cos(2x-3分之4派)+2cos^2x (1)求fx最大值时x的集合(2)已知三角形abc中,ABC的对边分别为abc.若f(B+C)=2分之3.b+c=2求a的最小值.

优质解答：

f(x)=cos(2x-4π/3)+2cos^2x

=cos(2x-4π/3)+cos2x+1

=2cos(2x-2π/3)cos2π/3+1

=1-√3cos(2x-2π/3)

1.

当 cos(2x-2π/3)=-1,2x-2π/3=2kπ+π,x=kπ+5π/6

f(x)最大=1+√3

2.

B+C=π-A

f(B+C)=1-√3cos(2(π-A)-2π/3)=3/2

cos(2A+2π/3)=-1/(2√3)

cos2Acos2π/3-sin2Asin2π/3=-√3/6

cos2A+√3sin2A=√3/3

√3cos2A+3sin2A=1

√3cos^2A-√3sin^2A+6sinAcosA=sin^2A+cos^2A

(√3-1)cos^2A+6cosAsinA-(√3+1)sin^2A=0

(√3+1)tan^A-6tanA-(√3-1)=0

△=(-6)^2-4(√3+1)[-(√3-1)]=36+4*2=44

tanA=[6±4√11)/[2(√3+1)]=[3±4√11)/(√3+1)

a^2=b^2+c^2-2bccosA

=(b+c)^2-2bc-2bccosA

=4-2(1+cosA)bc

b+c=2>=2√(bc),bc=1

a^2>=4-2(1+cosA)=2(1-cosA)

a 最小=√[2(1-cosA)]

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