<s>[INST] How many triangles have sides $a$, $a+1$, $a+2$ and angles $x$ and $2x$? 
How many triangles can I draw if its arms are $a$, $a+1$, and $a+2$, and two angles are $x$ and $2x$?


I think there are only two solutions to this problem. A triangle with arms $1$, $2$, $3$ and $2$, $3$, $4$. But I am not sure. 
 [/INST] This can be done with $a = 1, \frac{1}{2}(1 + \sqrt{13})$ and $4$ only. Not for $a = 2$. It's a simple matter of putting a set of equations into a symbolic calculator and letting it chew through them.
If the angles are $x, y, z$. Then the Cosine Law says, for instance,
$$ a^2 = (a+1)^2 + (a+2)^2 - 2(a+1)(a+2) \cos x $$
and we also have,
\begin{align*}
0 \le x &\le \pi \\
0 \le y &\le \pi \\
0 \le z &\le \pi \\
x + y + z &= \pi
\end{align*}
Then put in the various conditions like $x = 2z$ or $y = 2x$.
For example, in Mathematica,
Reduce[{a^2 == (a + 1)^2 + (a + 2)^2 - 2 ((a + 1) (a + 2)) Cos[x],
  (a + 1)^2 == a^2 + (a + 2)^2 - 2 (a (a + 2)) Cos[y],
  (a + 2)^2 == a^2 + (a + 1)^2 - 2 (a (a + 1)) Cos[2 x],
  3 x + y == Pi, 0 <= x,  0 <= y, 1 <= a}]

Tells you that $a = 4$ is a solution when $z = 2x$.
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