f(x)=sin(2x+a)+2√3*[cos(2x+a)+1]/2-√3
=sin(2x+a)+√3cos(2x+a)
=2sin(2x+a+z)
其中tanz=√3/1=tan(π/3)
所以f(x)=2sin(2x+a+π/3)
T=2π/2=π
f(-x)=2sin(-2x+a+π/3)=f(x)=2sin(2x+a+π/3)
sin相等则
-2x+a+π/3=2kπ+2x+a+π/3
或-2x+a+π/3=2kπ+π-(2x+a+π/3)
-2x+a+π/3=2kπ+2x+a+π/3
x=-kπ/2
此时于a无关,且只有特定的x才成立,不是恒等式
-2x+a+π/3=2kπ+π-(2x+a+π/3)
a=kπ+π/6
0<=a<=π
所以a=π/6