解由f（x）
=a*b=sin（2x+θ）+√3cos（2x+θ）
=2[1/2sin（2x+θ）+√3/2cos（2x+θ）]
=2sin（2x+θ+π/3)
又由函数f(x)=ab为偶函数,且θ∈[0,π]
则θ+π/3=π/2
即θ=π/6
故f（x）=2sin（2x+π/6+π/3)
=2sin（2x+π/2)
=2cos2x
2由x∈(0,π/2)
则2x∈(0,π)
又由f（x）=2cos2x=1
即cos2x=1/2
故2x=π/3
即x=π/6.