Question

已知函数fx)=根号3/2 Sin2x -Cos^2 x -1/2

Answer 1

f(x)=√3/2sin2x-(1+cos2x)/2-1/2

=√3/2sin2x-1/2cos2x-1

=sin(2x-π/6)-1

则最小正周期T=2π/2=π

f(c)=0,得sin(2c-π/6)=1,即2c-π/6=π/2,得：c=π/3

sinB/b=sinA/a

由sinB=2sinA,得：2sinA/b=sinA/a,得：b=2a

由余弦定理：c²=a²+b²-2abcosC

即3=a²+4a²-2a²

得a=1

故b=2