Polarization of an EM wave passing through another medium

[fluidistic]

Homework Statement

2 media whose refractive indices are 1 and n respectively are separated by a flat interface.
An EM plane wave goes from medium 1 to medium 2 with a polarization vector making an angle of 45° with the plane of incidence.
Determine the incidence angle for which the reflected wave has a polarization vector making an angle of 90° with respect to the plane of incidence. Show that in this case the polarization vector of the refracted wave forms an angle with the normal of the plane of incidence such that $\tan \theta= \left [ \frac{1}{2} \left ( n+ \frac{1}{n} \right ) \right ]$.

Homework Equations

Fresnel's.

The Attempt at a Solution

At first I was lost, I didn't know which equations to use. All the equations I've seen seem to be dealing with the magnitude of the amplitude of the electric field of the EM waves and not their phase. But I think I can use a Fresnel equation.
In the case that the incident wave has a polarization vector making an angle of 45° with respect to the incident plane, it means that there's a component alongside the plane of incidence and another along an orthogonal line to it. In other words if the plane wave is linearly polarized which I believe is ok, and if the plane of incidence is the y-z plane then I can write $\vec E =E_0(\hat x + \hat z) e^{i(ky - \omega t)}$.
Now for the reflected wave to make an angle of 90° with respect to the incident plane, it means it should have no $\hat y$ component, so using a Fresnel equation (formula #3 at https://en.wikipedia.org/wiki/Fresnel_equations#Formulas), the condition is $r_p=0$ so that $n \cos \theta_i = \cos \theta _t$ (1). Alsong with Snell's law $\sin \theta_i = n \sin \theta_t$ (2), I've got a system of 2 equations with 2 unknowns. My problem is that it's way too horrible to solve by hand.
I get that $\theta_i=\arctan [n^2 \tan [\arcsin (\frac{1}{n} \sin \theta_i)]]$. Of course I'm stuck on solving for $\theta_i$. Thus I guess I goofed somewhere but I really don't see it. Or maybe there's a huge simplification I've not seen...
Thanks for any help!

 

[BigJDubs]

A:You have to solve $\tan \theta = \left [ \frac{1}{2} \left ( n+ \frac{1}{n} \right ) \right ]$ for $\theta$.Substitute $\sin \theta = x$ and then solve $\sin \theta_i = \frac{1}{n}x$ for $x$. Then $\tan \theta = \left [ \frac{1}{2} \left ( n+ \frac{1}{n} \right ) \right ] \Rightarrow \tan \theta = \frac{1}{2} \left ( n+ \frac{1}{n} \right ) \frac{1}{x} \Rightarrow \theta = \arctan \left ( \frac{1}{2} \left ( n+ \frac{1}{n} \right ) \frac{1}{x} \right )$.