y=3sin^2x-4sinx+1=3(sin^2x-4/3sinx)+1=3(sinx-2/3)^2-1/3因为x属于[π/3,π]所以sinx属于[0,1]所以sinx=2/3时有最小值-1/3sinx=0时有最大值1从而函数y=3sin^2x-4sinx+1,x属于[π/3,π]的值域[-1/3,1]...