已知函数f(X)=2sin(x+π/6)-2cosx 若0

f（x）=2(sinxcosπ/6+cosxsinπ/6)-2cosx
=2(根号3/2*sinx+1/2*cosx)-2cosx
=根号3*sinx-cosx
=2(cosπ/6sinx-sinπ/6cosx)
=2sin(x-π/6),x∈[0,π/2].
00
所以2x-π/3∈(π/3,π/2)
sin(2x-π/3)=24/25
∴
cos(2x-π/6)=cos（2x-π/3+π/6）=cos(2x-π/3)cosπ/6-sin(2x-π/3)sinπ/6
=7/25*根号3/2-24/25*1/2
=7根号3-24/50

f(X)=2sin(x+π/6)-2cosx =sinx+根号3cosx-2cosx=sinx+（根号3-2）cosx
=5/6 再根据sin^2x+cos^2x=1 算出sinx cosx cos(2x-π/6)=2根号3cos2x+1/2sin2x