令sin(x-π/6)=t则∵ x∈[0,π/3]∴ x-π/6∈[-π/6,π/6]∴ sin(x-π/6)∈[-1/2,1/2]且 cos(2x-π/3)=cos[2(x-π/6)]=1-2sin²(x-π/6)=1-2t²y=cos（2x-∏/3）+2sin（x-∏/6）=1-2t²+2t=-2(t-1/2)
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