|2a-b|=|(2cosx-√3,2sinx+1)|
=√[(2cosx-√3)^2+(2sinx+1)^2]
=2√(2+sinx-√3cosx)
=2√[2(1+1/2*sinx-√3/2*cosx)]
=2√[2(1+sin(x-π/3))]
=2√[2(1+cos(5π/6-x))]
=2√[2*2cos^2 (5π/12-x/2)]
=4|cos(5π/12-x/2)|
x∈[0,π]
故-π/12≤5π/12-x/2≤5π/12
cos(5π/12)=(√6-√2)/4≤cos(5π/12-x/2)≤1
故√6-√2≤|2a-b|=4|cos(5π/12-x/2)|≤4
故m>4
即使是x∈(0,π),因x可取到5π/6,故答案也是m>4.而不是m≥4!