How can you solve this trigonometric equation algebraically? Solve for $0\leq\alpha\leq\pi$,
$$\left(\frac{\left(r-t\right)\cos\left(\alpha\right)+\left(r-t\right)\sin^{2}\left(\alpha\right)+t}{r}\right)^{2}+\left(\frac{4\left(t-r\right)\sin\left(\frac{\alpha}{2}\right)\cos^3\left(\frac{\alpha}{2}\right)}{t}\right)^{2}=1.$$
I've been playing with this for a while, and apparently the solution can be described simply by
\begin{align}
\alpha=2\arccos\left(\sqrt{\frac{t}{t+r}}\right)
\end{align}
and graphing the function with $\alpha$ on the $y$-axis, $t$ on the $x$-axis, and $r$ being an arbitrary constant, it does appear true. I'm not sure if this value of $\alpha$ holds true for all $r$ and $t$, though.
$\alpha$ on the $y$-axis and $t$ on the $x$-axis." />
In the above graph, the red curve is $\alpha=2\arccos\left(\sqrt{\frac{t}{t+r}}\right)$ and the purple is the original equation.
 A: Let
$$x := \left(\frac{\left(r-t\right)\cos\left(\alpha\right)+\left(r-t\right)\sin^{2}\left(\alpha\right)+t}{r}\right)^{2}+\\
\left(\frac{4\left(t-r\right)\sin\left(\frac{\alpha}{2}\right)\cos^3\left(\frac{\alpha}{2}\right)}{t}\right)^{2}-1. \tag{1}$$
Wanted: find the values of $\,\alpha\,$ when $\,x=0.$
There are a few ways to solve this problem.
One way with computer technology is to use the
substitution
$$\alpha=\frac{\log(A)}i,\;\;
  \sin(\alpha)=\frac{A-\frac1A}{2i},\;\;
 \cos(\alpha)=\frac{A+\frac1A}2 \tag{2}$$
in equation $(1)$ and factor the expression
using a CAS (Computer Algebra System) to get
$$ x = (1 - A)^2 (t - r) u^2 v/(4 A^2 r\, t)^2 \tag{3}$$ where
$$ u = r + 2 A r + A^2 r - t - A^2 t,\tag{4}$$
$$ v = r + 2 A r + A^2 r + t - 2 A t + A^2 t. \tag{5}$$
The factorization in equation $(3)$ implies that there
are three solutions for $\alpha.\,$ The first is
$\,\alpha=0\,$ when $\,A=1.\,$
The other two are for $\,u=0\,$ and $\,v=0.\,$
Note that
$$ \frac{2r}t-
\frac{\cos(\alpha)}{\cos(\alpha/2)^2}
 = \frac{2u}{t(1+A)^2} \tag{6}$$
and
$$ \cos(\alpha/2)^2 - \frac{t}{t+r} = \frac{v}{4(r+t)A}. \tag{7}$$
Thus $\,u=0\,$ when $$ \frac{2r}t = \frac{\cos(\alpha)}{\cos(\alpha/2)^2} \;\;\text{ or }\;\;
\frac{t}{2(t-r)} = \cos(\alpha/2)^2 \tag{8}$$
is one solution and $\,v=0\,$ when
$$ \frac{t}{t+r} = \cos(\alpha/2)^2 \quad
\text{ or } \quad\frac{r}{t} = \tan(\alpha/2)^2\tag{9}$$ is another
which agrees with your solution
$$ \alpha=2\arccos\left(\sqrt{\frac{t}{t+r}}\right). \tag{10}$$
