Question

设函数f（x）=cos（x+2π/3)+2(cosx/2)^2,x∈R

Answer 1

1.

f(x)=cos(x+2π/3)+2(cosx/2)^2

=cosxcos(2π/3)-sinxsin(2π/3)+cosx+1

=(-1/2)*cosx-(√3/2)*sinx+cosx+1

=(1/2)*cosx-(√3/2)*sinx+1

=cosxcos(π/3)-sinxsin(π/3)+1

=cos(x+π/3)+1

所以值域是【0,2】

2.

f(B)=1

所以cos(B+π/3)+1=1

即cos(B+π/3)=0

所以B+π/3=kπ+π/2(k∈Z)

又B是三角形内角

那么0＜B＜π

所以B=π/6

由余弦定理有cosB=(a^2+c^2-b^2)/2ac=(a^2+3-1)/(2a√3)=cos(π/6)=√3/2

所以a^2-3a+2=0

所以a=1或a=2

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