f(x)=4sinx(sinx-根号3cosx)-3,x∈[0,π/2],求f（x）值域

f(x)=4sinx*2sin(x-π/3)-3=-4【cos(x+x-π/3)-cos(x-x+π/3)】-3
=-4cos(2x-π/3)-1
所以当x∈[0,π/2]时,2x-π/3∈【-π/3,2π/3】
所以当x=π/2时,f(x)max=1
当x=π/6时,f(x)min=-5
所以f(x)的值域为【-5,1】