f(x)=sin(2x+π/6)+sin(2x-π/6)+cos(2x)+a
=sin(2x)cos(π/6)+cos(2x)sin(π/6)+sin(2x)cos(π/6)-cos(2x)sin(π/6)+cos(2x)+a
=2sin(2x)cos(π/6)+cos(2x)+a
=√3sin(2x)+cos(2x)+a
=2sin(2x+π/6)+a
（1）f(x)的最小正周期为π
（2）x=π/6时,f(x)max=f(π/6)=2+a
x=-π/3时,f(x)min=f(-π/3)=-2+a
∴f(x)max+f(x)min=2a=3
∴a=3/2
（3）f(x)=2sin(2x+π/6)+3/2
∴单调递减区间[kπ+π/6,kπ+2π/3]