Two polynomial equations in 2 unknowns I have a problem with this system of two polynomial equations in two unknowns.
$$
\begin{cases}
X_p=R_1 \cos\theta_1-R_3 \cos\theta_2-R \\[4px]
Y_p=R_1 \sin\theta_1-R_3 \sin\theta_2-R_2
\end{cases}
$$
where $X_p$, $Y_p$, $R_1$, $R_3$, $R_2$ and $R$ are known. The problem is knowing $\theta_1$ and $\theta_2$. I initially used trigonometric identities: $\cos\theta_1= (1-t_1^2)/(1+t_1^2 )$ $\sin\theta_1= (2t_1)/(1+t_1^2 )$
 A: Let 


*

*$A = (0,0)$

*$B = (c\cos\theta,c\sin\theta) \stackrel{def}{=} (X_p + R, Y_p + R_2)$

*$C = (R_1\cos\theta_1,R_1\sin\theta_1)$
and identity $\mathbb{R}^2$ with the Euclidean plane. The equation at hand is equivalent to
$$\begin{align}\overrightarrow{BC} = \overrightarrow{AC} - \overrightarrow{AB}
&= ( R_1\cos\theta_1 - (X_p + R), R_1\sin\theta_1 - (Y_p + R_2) )\\
&=  (R_3\cos\theta_2,R_3\sin\theta_2)
\end{align}
$$
The problem becomes finding a point $C$ such that
$$b \stackrel{def}{=} |AC| = R_1
\quad\text{ and }\quad a \stackrel{def}{=} |BC| = R_3$$
Apply cosine rule to $\triangle ABC$ at angle $A$ and notice
$\angle BAC = \theta_1 - \theta \pmod{2\pi}$, we obtain
$$\cos(\theta_1 - \theta) = \frac{b^2+c^2 - a^2}{2bc}\\
\implies
\theta_1 = \theta \pm \cos^{-1}\left(\frac{R_1^2 + c^2 - R_3^2}{2R_1 c}\right) \pmod{2\pi}$$
If $|R_1 - R_3| < c < R_1 + R_3$, there are two solutions of $\theta_1$ and hence two possible positions of $C$. Once you get $C$, extracting $\theta_2$ from $\overrightarrow{BC}$ will be routine. 
Notice $\angle CBA = (\pi + \theta) - \theta_3 \pmod{2\pi}$, there is a similar formula for $\theta_3$.
$$\theta_3 = \pi + \theta \pm \cos^{-1}\left(\frac{R_3^2 + c^2 - R_1^2}{2R_3 c}\right) \pmod{2\pi}$$
For $|R_1 - R_3| < c < R_1 + R_3$, there are again two solutions for $\theta_3$.
However, these two solutions of $\theta_3$ are not independent from those of $\theta_1$. It is not immediately clear to me how to pair up solutions of $\theta_3$ with $\theta_1$. That's why I suggest you once $\theta_1$ is known, extract $\theta_3$ out from $\overrightarrow{BC}$ instead.
