f(x)=(1-cos2x)/2+√3sin2x-(1+cos2x)/2=√3sin2x-cos2x=2(sin2x*cosπ/6-cos2x*sinπ/6)=2(sin2x-π/6)1)当2x-π/6=2kπ+π/2时，f(x)取最大值为2此时x=kπ+π/3, 这里k为任意整数2）函数的单调增区间为： 2kπ-π/2=<2x-π/6<=2kπ+π/2即kπ-π/6=所以在[0,π]内，单调增区间为[0, π/3]U[5π/6, π]
