已知函数f（x）=2sinxcosx+根号3（2cos^x-1）（1）将函数f（x）化为Asin（wx+f）（w大于0,

(1)f(x)=2sinxcosx+√3（cos^2x-1）=sin2x+√3cos2x=2sin(2x+π/3)
(2)由-π/2+2kπ≤2x+π/3≤π/2+2kπ,得,-5π/12+kπ≤x≤π/12+kπ (k是整数)
由π/2+2kπ≤2x+π/3≤3π/2+2kπ,得,π/12+kπ≤x≤7π/12+kπ （k是整数）
故f(x)的单调递增区间是{x|-5π/12+kπ≤x≤π/12+kπ }
单调递减区间是{x|π/12+kπ≤x≤7π/12+kπ} (k是整数)