When x is set to {x | x = 2kπ+π/2, k ∈ z}, the maximum value is 2.

When x is set to {x | x = 2kπ-π/2, k ∈ z}, the minimum value is -2.

②y=2-cosx/3，X∈R

When x/3=2kπ, k∈Z, that is, x=6kπ, k∈Z,

Cosx/3 gets the maximum value 1, and y gets the minimum value 1.

∴y gets the minimum value of 1, at which time the x set {x | x = 6kπ, k ∈ z}

When x/3=2kπ+√, k∈Z, that is, x=6kπ+2π, k∈Z,

Cosx/3 gets the minimum value-1, and y gets the maximum value 3.

The minimum value of ∴y is 3, and the x set {x | x = 6kπ+2π, k ∈ z}

③

From 2kπ+π/2 ≤ 2x+π≤2kπ+3π/2.

Get 2kπ+π/4≤2X≤2kπ+5π/4, k ∈ z.

∴kπ+π/8≤2X≤kπ+5π/8,k∈Z

∴ monotone decreasing interval of function

[kπ+π/8，kπ+5π/8]，k∈Z