（1）f(x)=2sin(x+π/6)-2cosx=3^(1/2)sinx-cosx=2sin(x-π/6)
所以,其实单调增区间应满足,2kπ-π/2≤x-π/6≤2kπ+π/2
求得2kπ-π/3≤x≤2kπ+2π/3
（2）f(x)=2sin(x-π/6)=6/5,则sin(x-π/6)=3/5
cos(2x-π/3）=cos2(x-π/6) =1-2sin^2[(x-π/6)]=1-2*(3/5)^2=1-18/25=7/25