你这里是不是少打个条件x∈[0,π/2]啊f(x)=cos^4x-2sinxcosx-sin^4x =（cosx)^2-(sinx)^2-2sinxcosx=cos2x-sin2x=(√2)cos(2x+π/4),(1)f（x)的最小正周期是π.（2）x∈[0,π/2],∴2x+π/4∈[π/4,5π/4],∴f(x)的最...没有那个条件f(x)的最大值就是当2X+π/4=1即2Kπ+π/2=2X+π/42Kπ+π/2=2X+π/42Kπ+π/2-π/4=2X2X=2Kπ+π/4X=Kπ+8/π(K∈Z)此时f(x)max=√2f(x)的最小值就是当2X+π/4=-1即2Kπ+π=2X+π/42Kπ+π/2-π/4=2X2X=2Kπ+π/4X=Kπ+π/8(K∈Z)此时f(x)min=-√2不懂的欢迎追问,如有帮助请采纳，谢谢!