Question

已知函数f(x)=√3sin^2x-sinxcosx,x属于【π/2,π】.求方程f(x)=0的根

Answer 1

f(x)=√3sin^2x-sinxcosx
=√3(1-cos2x)/2-1/2*sin2x
=√3/2-√3/2cos2x-1/2sin2x
=-(1/2sin2x+√3/2cos2x)+√3/2
=-sin(2x+π/3)+√3/2
f(x)=0
-sin(2x+π/3)+√3/2=0
sin(2x+π/3)=√3/2
x属于【π/2,π】
∴2x∈[π,2π]
2x+π/3∈[4π/3,7π/3]
∴2x+π/3=7π/3
2x=2π
x=π
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Answer 2

f(x)=sinx(√3sinx-cosx)
=sinx[2sin(x-π/6)]
sinx[2sin(x-π/6)]=0
因为，π/3≤(x-π/6)≤5π/6,所以，
sin(x-π/6)≠0
sinx=0==>x=π