函数f(x)=sin(2x-兀/6)+2sin^2(x-兀/12) =sin(2x-兀/6)+1-cos(2x-兀/6)
=√2sin(2x-5兀/12)+1
∴f(x)的最小正周期为兀
2x-5兀/12=2k兀+兀/2==>x=k兀+11兀/24时,取极大值1+√2
函数f(x)=sin(2x-兀/6)+2sin^2(x-兀/12) =sin(2x-兀/6)+1-cos(2x-兀/6)
=√2sin(2x-5兀/12)+1
∴f(x)的最小正周期为兀
2x-5兀/12=2k兀+兀/2==>x=k兀+11兀/24时,取极大值1+√2