Question

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)=0

sinB=2sinA,求a，b的值

Answer 1

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=0
2sin(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=2sinA，所以B=2A=π-C-A=2π/3-A
所以sin(2π/3-A)=2sinA
展开得sin2π/3cosA-cos2π/3sinA=2sinA
即3/2sinA=根号3/2cosA
所以tanA=sinA/cosA=根号3/3
所以A=π/6
又因为C=π/3，所以B=π/2
因为sinB=2sinA,由正弦得b=2a
所以a^2+c^2=b^2=(2a)^2
解得a=1, 所以b=2a=2

采纳吧= =，打了好久，认认真真自己做的好辛苦

Answer 2

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<C<π/2
∴0<2C<π
∴-π/6<2C-π/6<5π/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