设m=1/2x+π/4
则函数y=sinm的单调递增区间为[-π/2+2kπ,π/2+2kπ]其中k∈Z
将m=1/2x+π/4代入得：
-π/2+2kπ＜1/2x+π/4＜π/2+2kπ
化简得-3/2π+4kπ＜x＜π/2+4kπ 其中k∈Z
∵x∈[-2π,2π],∴k=0.
∴-3/2π＜x＜π/2
即f(x)在[-2π,2π]的单调增区间为[-3/2π,π/2].