I'm given that $$\vec E =\frac{\mu_0p_0\omega^2}{4\pi r}\Big[\cos u\Big(\hat x-\frac{x}{r}\hat r\Big)+\sin u \Big(\hat y-\frac{y}{r}\hat r\Big)\Big]$$ implies $$-\frac{\mu_0p_0\omega^2}{4\pi r}\Big[\cos u\hat x\times \hat r + \sin u \hat y \times \hat r\Big]=\frac{1}{c}\hat r \times \vec E,$$ but I don't follow how to get from the former to the latter.
Could someone connect the dots for me?
You get the second equation by taking the cross product of the first one with $\frac{1}{c} \hat{r}$ on the left and remembering that $\vec{x} \times \vec{x} = \vec{0}$, so in particular $\hat{r} \times \hat{r} = \vec{0}$; and also that $\hat{r}\times\hat{x} = - \hat{x} \times \hat{r}$.