Algebraic solution for the value of $x$. I solved this problem the fifteen years ago without numerically solving equations of degree 4, I was happy in a substitution that I avoid directly attacking equations of degree 4. 

Today my nephew, who is an enthusiastic student of mathematics, proposes me the same problem. It occurs to me that I am a very rusted to algebraic substitutions exhaustively. I tried for about 3 hours. How disappointed do not want my nephew to his attempts (and not me) ask for help to the ME community. And of course I will give all the credits to ME. 
My attempt.

Note that
$
\cos \alpha = \frac{1}{x}$, $\cos \alpha = \frac{y}{1}$, $\cos \alpha = \frac{y+1}{x+1}$. Then we have 
$$
xy=1.
$$
By Pythagoras Theorem we have $(y+1)^2+1^2=(x+1)^2 \Longleftrightarrow y^2+2y+1+1=x^2+2x+1$, i.e.
$$
x^2-y^2+2(x-y)-1=0.
$$
Update [ July 26 2016 ] I remember the time I solved this question, I tried something like
\begin{align}
x^2-y^2+2(x-y)-1=0
&
\Longleftrightarrow
(x-y)[x+2+y]=1
\\
&
\Longleftrightarrow
(x-y)[x+2\cdot 1+y]=1
\\
&
\Longleftrightarrow
(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})[x+2\sqrt{x}\sqrt{y}+y]=1
\\
&
\Longleftrightarrow
(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})^3=1
\\
&
\Longleftrightarrow
(\sqrt{x}+\sqrt{y})^3=\frac{1}{(\sqrt{x}-\sqrt{y})}
\\
&
\Longleftrightarrow
(\sqrt{x}+\sqrt{y})^3=\frac{(\sqrt{x}+\sqrt{y})}{(x-y)}
\\
&
\Longleftrightarrow
(\sqrt{x}+\sqrt{y})^2=\frac{1}{(x-y)}
\\
&
\Longleftrightarrow
(x+y+2)=\frac{1}{(x-y)}
\end{align}
We then have two ways to tackle the problem:
$$
\left\{\begin{array}{rl}\sqrt{x}\sqrt{y}=&1
\\
 (\sqrt{x}+\sqrt{y})^3=&\frac{1}{(\sqrt{x}-\sqrt{y})} \end{array}\right.
\quad\mbox{ or } \quad 
\left\{\begin{array}{rl}xy=&1
\\ 
(x+y+2)=&\frac{1}{x-y} \end{array}\right.
$$
 A: From $y = 1/x,$ then multiplying by $x^2,$ i got
$$ x^4 + 2 x^3 - x^2 - 2 x - 1. $$
This looks bad. However, set
$$ x = t - \frac{1}{2} $$
and you get rid of the cubic term, always worth a try. I was pleased to discover that the linear term also vanished, giving
$$ t^4 - \frac{5}{2} t^2 - \frac{7}{16}, $$
and you can solve for $t^2$ with the Quadratic Formula. I get
$$ t^2 = \frac{5 \pm \sqrt {32}}{4} $$
with two pure imaginary roots, a real negative, and a real positive for $t$ itself. Then $x$ is that minus 1/2. I get $1.132241883$ as $x.$
One good habit is to simply draw a graph of the function. I do them by hand with a calculator to find points. I have appended a good online graph. Notice that the graph appears to be symmetric across the vertical line $x = -\frac{1}{2}.$ We could confirm this by taking $ f(x) =  x^4 + 2 x^3 - x^2 - 2 x - 1 $ and then checking whether $f(-1-x) = f(x)??$ In turn, this confirmation would tell us that the translation I tried would, in fact, give a graph symmetric across the y axis, meaning all even exponents. 
Calculus ideas that are, at least, consistent with the symmetry notion include $$ f'(x) = 2 (2x+1) \left(x^2 + x -1 \right)  $$ and
$$ f''(x) = 2 \left(6 x^2 + 6 x - 1 \right), $$
so that $x=-1/2$ gives a local maximum, while the inflection points are  symmetric around $x = -1/2$ by the quadratic formula.
Huh. Turns out the local minima really are along $y = -2,$ since
$$ f(x) + 2 = \left( x^2 + x - 1 \right)^2. $$
Go Figure.
=========================
parisize = 4000000, primelimit = 500509
? x = -1 - w
%1 = -w - 1
? x^4 + 2 * x^3 - x^2 - 2 * x - 1
%2 = w^4 + 2*w^3 - w^2 - 2*w - 1
? 

==================================
parisize = 4000000, primelimit = 500509
? factor( x^4 + 2 * x^3 - x^2 - 2 * x - 1 )
%1 = 
[x^4 + 2*x^3 - x^2 - 2*x - 1 1]

? x = t - (1/2)
%2 = t - 1/2
? p = x^4 + 2 * x^3 - x^2 - 2 * x - 1
%3 = t^4 - 5/2*t^2 - 7/16
? 

==============================================


