Calculate angle of inclination for a certain angle in a planet's orbit?
In short, no, you don't have to solve Kepler's equation. If you define a viewing longitude and you know the full set of Keplerian orbital elements, you can calculate if a planet will show a transit.

Let me first clear up some of the jargon, though. The term inclination is traditionally reserved for the angle between the orbital plane and the reference plane. So unless the entire orbit is wobbling (for instance due to precession) the inclination will be constant for all positions of the planet.

The angle that varies as the planet moves through the reference plane is the latitude. However, in this particular problem calculating the latitude isn't particularly useful. The more practical coordinate to look at is the height of the planet away from the plane, I'll call this $r_z$. If along your observers line of sight you find $r_z$ to be smaller than the solar radius, the planet will show a transit.

Working out a fully general elliptical orbit in 3 dimensions is somewhat tricky, so I'll start out with a simple Cartesian coordinate system aligned with the orbital plane. In subsequent steps I work my way back using a number of coordinate transformations.

First a list of the Kepler parameters and their meaning:


$\nu$ - true anomaly - The anglular position with respect to perigee in the orbital plane.
$\omega$ - argument of periapsis - The angular position of perigee with respect to the ascending node in the orbital plane.
$\lambda_{asc}$ - the longitude of ascending node in the invariable plane ($\Omega$ in the picture).
$i$ - inclination - the angle between the orbital plane and the invariable plane, such that $\lambda_{asc}$ gives the rotation axis. 




Now, in the orbital plane, the radius of the planet orbit is given as
$$
r(\nu) = \frac{a(1-e^2)}{1+e\cos(\nu)}
$$
If I define the x-axis to align with the ascending node, I can write the position vector as
$$
\vec{r}_{orb} = r(\nu) 
\begin{bmatrix} 
\cos(\omega+\nu) \\ 
\sin(\omega+\nu) \\ 
0 
\end{bmatrix}
$$

To move to a Cartesian coordinate system aligned with the invariable plane I have to apply a rotation of angle $i$ about the x-axis, so
$$
\vec{r} = R_x(i)\vec{r}_{orb}= r(\nu)
\begin{bmatrix} 
\cos(\omega+\nu) \\ 
\sin(\omega+\nu)\cos(i) \\ 
\sin(\omega+\nu)\sin(i)
\end{bmatrix}
$$
So now I have an expression for $r_z$, I just need to know how $\nu$ relates to the planet longitude in the invariable plane, $\lambda_{planet}$.

I can find the longitude of the planet relative to $\lambda_{asc}$ as
$$
\lambda'_{planet} = \arctan( \frac{r_y}{r_x} ) 
$$ 
such that the actual longitude is
$$
\lambda_{planet} = \lambda_{asc} + \lambda'_{planet}
$$
Inverting this last expression gives
$$
\nu(\lambda) = \arctan( \frac{\tan(\lambda - \lambda_{asc})}{\cos(i)} ) - \omega
$$

Now I have all the equations to answer the question:
If the planet is along the observer's line of sight, then is has a true anomaly 
of $\nu(\lambda_{obs})$, which gives a height above the plane
$$
r_z(\nu) = r(\nu)\sin(\omega+\nu)\sin(i)
$$
If $|r_{z}| &lt; R_{sun}$ the planet can be seen in transit.