6/π≤x≤2π/3?应该是π/6≤x≤2π/3吧?
如果是的话,

令y=2x-π/3,则：f(y)=2siny
此时：x=y/2-π/6,
当x∈[π/6,2π/3],有：y∈[0,π]
对于f(y)=2siny,当y∈[0,π]时,有：f(y)∈[0,1]
所以,对于f(x)=sin(2x-π/3),当x∈[π/6,2π/3]时,有：f(x)∈[0,1]
即所求值域是：0≤f(x)≤1.