f(x)=sin^4x-2根号3sinxcosx-cos^4x+1
=(sin^2x-cos^2x)(cos^2x+sin^2x)-2根号3sinxcosx+1
=-cos2x-根号3sin2x+1
= -2sin(2x+π/6)+1
当0<=x<=Pai/3时,有Pai/6<=2x+Pai/6<=5Pai/6
1/2<=sin(2x+Pai/6)<=1
故有-2+1<=f(x)<=-2*1/2+1
即有-1<=f(x)<=0
当2x+Pai/6=Pai/6,即X=0时有最大值是0
题目可不可以说的清楚点，有点看不懂
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