若x∈[-兀/3,2兀/3],求函数y=cos(x+兀/3)-sin^2(x+兀/3)+1的最大值和

网友回答

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)
当cost=-(1/2)时,方程具有最小值为ymin=-1/4
当cost=1时,方程具有最大值为ymax=1^2+1=2