f(x)=sin(2x+θ)+2√3[1+cos(2x+θ)]/2-√3
=sin(2x+θ)+√3cos(2x+θ)
=2sin(2x+θ+π/3)是偶函数
f(-x)=f(x)
2sin(-2x+θ+π/3)=2sin(2x+θ+π/3)
所以-2x+θ+π/3=2kπ+2x+θ+π/3或-2x+θ+π/3=2kπ+π-(2x+θ+π/3)
-2x+θ+π/3=2kπ+2x+θ+π/3
4x=-2kπ
不是恒等式
-2x+θ+π/3=2kπ+π-(2x+θ+π/3)
2θ=2kπ+π/3
θ=kπ+π/6
所以θ=π/6