求函数y=√2sin（2x-π）cos[2(x+π）]的单调区间

∵y＝－√2sin（π－2x）cos（2π＋2x）＝－√2sin2xcos2x＝－（√2/2）sin4x.
∴当2kπ－π/2＜4x≦2kπ＋π/2时,y单调递增,
　当2kπ＋π－π/2＜4x≦2kπ＋π＋π/2时,y单调递减.
由2kπ－π/2＜4x≦2kπ＋π/2,得：kπ/2－π/8＜x≦kπ/2＋π/8.
由2kπ＋π－π/2＜4x≦2kπ＋π＋π/2,得：kπ/2＋π/8＜x≦kπ/2＋3π/8.
∴原函数的单调递增区间是（kπ/2－π/8,kπ/2＋π/8］,
　单调递减区间是（kπ/2＋π/8,kπ/2＋3π/8］.其中k为整数.