函数y=2sin^2x+2cosx-3,x∈[-π/3,2π/3]的最小值

y=2sin^2x+2cosx-3
=2(1-cos²x)+2cosx-3
=-2cos²x+2cosx-1
=-2(cosx-1/2)²-1/2
x∈[-π/3,2π/3]
∴cosx∈[-1/2,1]
∴当cosx=-1/2时
y=-2*（-1）²-1/2
=-2-1/2
=-5/2
为最小值
如仍有疑惑,欢迎追问.祝：学习进步!