已知函数f(x)=sin(2x- π/6)+2cos²x-1 求单调增区间

f(x)=sin(2x- π/6)+2cos²x-1 =sin2xcosπ/6-cos2xsinπ/6+cos2x=√3/2 sin2x+1/2cos2x=sin(2x+π/6)由2kπ-π/2≤2x+π/6≤2kπ+π/2,k∈z得 kπ-π/3 ≤x≤kπ+π/6,k∈z所以单调递增区间为[kπ-π/3 ,kπ+π/6],k∈z