已知函数f（x）=sin^4(wx)+cos^4ωx的相邻对称轴之间的距离为π/2 （1）求正数ω的值

1.f（x）=sin^4(wx)+cos^4ωx=(sin²wx)²+(cos²wx)²=(1-cos2wx)²/4+(1+cos2wx)²/4
=(1+1+2cos²2wx)/4=1/2+1/2cos²2wx=1/2+1/2(1+cos4wx)/2=3/4+1/4cos4wx
相邻对称轴之间的距离为π/2,所以T/2=π/2,所以2π/4w=π,所以w=1/2
2.g(x)=2[3/4+1/4cos2x]+sin²(x+π/6)=3/2+1/2cos2x+[1-cos(2x+π/3)]/2
=2+1/2cos2x-1/2(cos2xcosπ/3-si2xsinπ/3)
=2+1/2cos2x-1/4cos2x+√3/4sin2x
=2+√3/4sin2x+1/4cos2x=2+1/2sin(2x+π/6)
所以,最大值为2+1/2=5/2,此时2x+π/6=2kπ+π/2,所以x=kπ+π/6