已知函数f(x)=根号3sin2x-2cos^2x-1,x∈R,在△ABC中,A,B,C的对边分别为a b c 已知 c=根号3,f(c)=0sinB=2sinA,求a,b的值

f(x)=√3sin2x-2cos^2x-1,x∈R
=√3sin2x-(1+cos2x)-1
=√3sin2x-cos2x-2
=2(√3/2sin2x-1/2cos2x)-2
=2sin(2x-π/6)-2
∵f(C)=0
∴2sin(2C-π/6)-2=0
sin(2C-π/6)=1
∵0∴0∴-π/6∴2C-π/6=π/2
∴C=π/3,B=π-C-A=2π/3-A
∵sinB=2sinA
∴sin(2π/3-A)=2sinA
∴sin2π/3cosA-cos2π/3sinA=2sinA
∴3/2sinA=√3/2cosA
∴tanA=sinA/cosA=√3/3
∴A=π/6
∴B=π/2

又c=√3,
根据正弦定理：
a/sinA=b/sinB=c/sinC
∴a=csinA/sinC=1
b=√(a²+c²)=2

f(x)=根号3sin2x-(2cosx^2-1)-1-1=根号3sin2x-cos2x-2=2(根号3/2sin2x-1/2cos2x)-2=2sin(2x-π/6)-2因为f(C)=0,所以f(C)=2sin(2C-π/6)-2=02sin(2C-π/6)=2,即sin(2C-π/6)=1所以2C-π/6=π/2解得C=π/3因为sinB=2si...