采用换元法
令cosx=t,当x∈[-π/4,0]时,t=cosx单调递增；当x∈[0,π/3]时,t=cosx单调递减
所以t∈[1/2,1]
此时y=cos^2x-3cosx+1(x∈[-π/4,π/3])可转化为：y=t^2-3t+1,t∈[1/2,1]
因为y=t^2-3t+1在[1/2,1]上递减（y=t^2-3t+1在(-∞,3/2]上递减,在[3/2,+∞)递增）
所以其值域为[-1,-1/2]