已知函数f(x)=√3sin(x/4)cos(x/4)+cos²(x/4) 1 若f(x)=1，求

已知函数f(x)=√3sin(x/4)cos(x/4)+cos²(x/4) 1 若f(x)=1，求

f(x)=√3sin(x/2)/2+cos(x/2)/2+1/2
=sin(x/2+π/6)+1/2
1）f(x)=1，sin(x/2+π/6)=1/2
cos(2π/3-x)=cos[π-(x+π/3)]=-cos(x+π/3)=2[sin(x/2+π/6)]^2-1=-1/2
2）用余弦公式，cosC=(a^2+b^2-c^2)/2ab
代入等式，两边同时乘以b
a^2+b^2-c^2+bc=2b^2
即a^2=b^2+c^2-bc
对比cosA的余弦公式，A=π/3
在锐角三角形中，B的范围是(π/6,π/2)
f(2B)=sin(B+π/6)+1/2，取值范围((√3+1)/2,3/2)

追问第二题有没追答等等

(1)f(x)=√3/2*sin(x/2)+1/2*[cos(x/2)+1]
=√3/2*sin(x/2)+1/2*cos(x/2)+1/2
=sin(x/2+π/6)+1/2
若f(x)=sin(x/2+π/6)+1/2=1，那么sin(x/2+π/6)=1/2
那么cos[2(x/2+π/6)]=1-2sin²(x/2+π/6)=1-2*1/4=1/2
即cos(x+π/3)=1/2
∴cos(2π/3-x)=cos(x-2π/3)=-cos(x+π/3)=-1/2
(2)2acosC+c=2b
正弦定理：2sinAcosC+sinC=2sinB
而sinB=sin[π-(A+C)]=sin(A+C)=sinAcosC+cosAsinC
∴2sinAcosC+sinC=2(sinAcosC+cosAsinC)
∴sinC=2cosAsinC
而C是锐角，∴sinC≠0，∴cosA=1/2
又A是锐角，∴A=π/3，那么B+C=2π/3
∴C=2π/3-B，而C是锐角，即0∴0<2π/3-B<π/2，∴π/6而B是锐角，∴π/6∴π/3∴√3/2∴f(2B)=sin(B+π/6)+1/2∈((1+√3)/2，3/2]

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