f(x)=sin(2x+π/3)+sin(2x-π/3)+2cos^2*x+a-1
=sin2xcosπ/3+cos2xsinπ/3+sin2xcosπ/3-cos2xsinπ/3+2cos^2x-1+a
=2sin2xcosπ/3+2cos^2x-1+a
=sin2x+cos2x+a
=√2sin(2x+π/4)+a
当sin(2x+π/4)=1时,取最大值,即√2+a=√2+1 a=1
(2)f(x)的对称中心就是 =√2sin(2x+π/4)与x轴的交点(因为图像向上平移了一个单位,所以对称中心的纵坐标都为1)
即√2sin(2x+π/4)=0解得x=kπ/2-π/8.
所以对称中心的坐标为(kπ/2-π/8,1)
(3)g(x)=f(x+3/8π)+2=√2sin(2(x+3/8π)+π/4)+1+2=√2sin(2x+π)+3
=-√2sin2x+3
求-√2sin2x的减区间就行了,也就是求√2sin2x的增区间
2kπ-π/2《2x《2kπ+π/2 ,解出来 kπ-π/4《x《kπ+π/4 所以g(x)的减区间为【kπ-π/4,kπ+π/4】