Wald's General Relativity, section 6.3 Page 144 I cannot understand how he reaches the conclusion in equation 6.3.36 and 6.3.37; even the terminology is somewhat confusing. This is a problem of bending of light under gravitational field.
This is what he mentioned in the page: we have, for Schwarzschild metric,
\begin{equation}
\frac{d\phi}{dr}=\frac{L}{r^2}\left[E^2-\frac{L^2}{r^3}(r-2M)\right]^{-1/2}.\tag{6.3.35}
\end{equation}
We wish to find $\Delta\phi$ of light ray in Schwarzschild geometry, traversing a path near the spherical star.
In order to not be captured, the impact parameter $b$ must be greater than the critical impact parameter $$b_c=3^{3/2}M.\tag{6.3.33}$$ In that case, the orbit of the light ray will have a "turning point" at the largest radius, $R_0$, for which $V(R_0)=E^2/2$, i.e. at the largest root of $$R_0^3-b^2(R_0-2M)=0,\tag{6.3.36}$$ which is
\begin{equation}
R_0=\frac{2b}{\sqrt{3}}\cos\left[\frac{1}{3}\cos^{-1}\left(-\frac{b_c}{b}\right)\right].\tag{6.3.37}
\end{equation}
I am lost at this point:


*

*What does he mean by "largest radius" $R_0$?

*How does he suddenly make the cosine conclusion?


I could understand the rest of this section but these two remarks are too obscure to me. Any help is greatly appreciated.
 A: Let's start with the Schwarzschild geodesic
$$
\frac{1}{2}\dot{r}^2 + V(r) = \frac{1}{2}E^2,
$$
with $V(r)$ the effective potential
$$
V(r) = \frac{L^2}{2r^3}(r-2M).
$$
Consider light rays that are deflected in the Schwarzschild geometry, i.e. photons that start at a large distance ($r$ large, $\dot{r}<0$), propagate until they reach a distance of closest approach at coordinate $r=R_0$ (where $\dot{r}=0$), and propagate outward again. Therefore, $R_0$ is a solution of the equation
$$
\frac{1}{2}E^2 = V(r) = \frac{L^2}{2r^3}(r-2M),
$$
or 
$$
r^3 -b^2(r-2M) =0,
$$
with $b=L/E$. If $b<b_c = 3^{3/2}M$ then this equation has no positive roots, which means that the photon will continue to spiral towards the centre. If $b>b_c$, the equation has two positive roots and one unphysical negative root. We're only interested in the largest of the positive roots, since we started at a very large radius ($r \gg R_0$) and the photons will never reach any distance smaller than $R_0$.
The solution
$$
R_0 = \frac{2b}{\sqrt{3}}\cos\left[\frac{1}{3}\cos^{-1}(-b_c/b)\right]
$$
can be verified by using $4\cos^3x =\cos 3x + 3\cos x$. One can also verify that it is the largest root, since $b>b_c$ and
$$
\cos^{-1}(-b_c/b) < \cos^{-1}(-1) = \pi\\
\cos\left[\frac{1}{3}\cos^{-1}(-b_c/b)\right] > \cos(\pi/3) = 1/2\\
R_0 > \frac{b}{\sqrt{3}} > \frac{b_c}{\sqrt{3}} = 3M,
$$
and $3M$ is the coordinate where $V(r)$ has a maximum. The other root (let's call it $R_1$) will lie between $2M< R_1 < 3M$. This root applies to photons that start at very small radii $(2M < r < R_1)$ and try to escape, but cannot move beyond the turning point $R_1$, after which they spiral towards the centre. 
A: I don't know why he uses the name "largest radius" but physically it is the radius of closest approach, which is the minimum distance to the center of the star at the point of reflection of the orbit. Since it is a point of reflection $\dot{r}=0$ in the equation 6.3.14 
$$
\frac{1}{2}\dot{r}^2 + V_{\text{eff}}(r) = \frac{1}{2}\dot{r}^2 + \frac{L^2}{2r^2} \left( 1 - \frac{2M}{r}\right) = \frac{E^2}{2}
$$
this gives $V_{\text{eff}} (R_o) = E^2/R$ which is
$$
\frac{L^2}{2R_o^2} \left( 1 - \frac{2M}{R_o}\right) = \frac{E^2}{2}
$$
which is just
\begin{align}
&R_o^3 - b^2 (R_o - 2M) = 0 \\
&R_o^3 - b^2 (R_o - \frac{2b_c}{3\sqrt{3}}) = 0 \\
&4x^3 - 3x =- \frac{b_c}{b}
\end{align}
where $x \equiv \frac{\sqrt{3}R_o}{2b}$. Now Wald wants to flex his algebraic muscles so remember a formula that says $4\text{cos}^3\theta - 3\text{cos }\theta = \text{cos }3\theta$ so defining $\text{cos }\theta\equiv x$ gives $\theta = \frac{1}{3}\text{cos}^{-1}(-b_c/b)$ or finally
$$
x = \text{cos }\left(\frac{1}{3}\text{cos}^{-1}\frac{-b_c}{b}\right)\qquad\Rightarrow\qquad R_o = \frac{2b}{\sqrt{3}} \text{cos }\left(\frac{1}{3}\text{cos}^{-1}\frac{-b_c}{b}\right)
$$
