已知函数f(x)=2sin(x+π/6)-2cosx,x∈【π/2,π】,求实数m使不等式3m^2-m-f(x)≥0恒成

f(x)=2sin(x+π/6)-2cosx =2(sinx cosπ/6 + cosx sin π/6 -cosx)=2(sinx cosπ/6 - cosx sin π/6)
=2sin (x- π/6)
x∈【π/2,π】,x- π/6∈【π/3,5π/6】,所以 f(x)=2sin (x- π/6) ∈【1,2】
3m^2-m-f(x)≥0 即 3m^2-m ≥ f(x） 也就是说要使m满足3m^2-m≥2 即3m^2-m -2 ≥ 0
解该不等式.可得（m-1）(3m+2) ≥ 0
所以m ≥ 1 或 -2/3 ≥ m
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