2向量a-向量b=2√3cosx--2sinx
=4(√3/2cosx-1/2sinx)
=4sin(π/3-x)=--4sin(x-π/3)
因为x属於[0,π],所以x-π/3属於[---π/3,2π/3]
sin(x-π/3)属於[---√3/2,1],--4sin(x-π/3)属於[-4,2√3]
所以|--4sin(x-π/3) |属於[0,4]
因为m>| 2a -b |恒成立,所以m>| 2a -b |的最大值
所以m>4
2向量a-向量b=2√3cosx--2sinx
=4(√3/2cosx-1/2sinx)
=4sin(π/3-x)=--4sin(x-π/3)
因为x属於[0,π],所以x-π/3属於[---π/3,2π/3]
sin(x-π/3)属於[---√3/2,1],--4sin(x-π/3)属於[-4,2√3]
所以|--4sin(x-π/3) |属於[0,4]
因为m>| 2a -b |恒成立,所以m>| 2a -b |的最大值
所以m>4