1、对f(x)求导,得f'(x)=2cos(-2x+π/6)*(-2)=4cos(-2x+π/6)对函数f'(x)=4cos(-2x+π/6),当-π/2+2kπ≤-2x+π/6≤π/2+2kπ时,f'(x)≤0,f(x)单调递减,此时-π/6-kπ≤x≤π/3-kπ当π/2+2kπ≤-2x+π/6≤π+2kπ时,...