解.f(x)=2sinx[1-cos(x+π/2)]+1-2sin²x=2sinx(1+sinx)+1-2sin²x=2sinx+1
(1)y=f(wx)=2sinwx+1
因在区间[-π/2,2π/3]上是增函数,所以最小正同期T=2π/w≥2(π/2+2π/3)
即0