由x∈[-兀/3,2兀/3]
令t=(x+兀/3)
∴t=(x+兀/3)∈[0,兀]
∴cost∈[-1,1]
∴y=cos(x+兀/3)-sin^2(x+兀/3)+1
=cost-sin^2t+1
=cost+(1-sin^2t)
=cos^2t+cost
=(cost+(1/2))^2-(1/4)

當(dāng)cost=-(1/2)時,方程具有最小值為ymin=-1/4
當(dāng)cost=1時,方程具有最大值為ymax=1^2+1=2