Solving the system $a_1\sin x_1+b_1\cos x_2=c_1$, $a_2\cos x_1+b_2\sin x_2=c_2$ Does the following system of equations have a closed-form solution? If so, how can I solve it?
$$\begin{align}
a_1\sin(x_1) + b_1\cos(x_2) &= c_1 \\
a_2\cos(x_1) + b_2\sin(x_2) &= c_2
\end{align}$$
 where $a_1$, $a_2$, $b_1$, $b_2$, $c_1$ and $c_2$ are constants. 
(I'm not looking for numerical solution.)
 A: I'll change notation to this:
$$\begin{align}
p_1 \cos x_1 + q_1 \sin x_2 &= r_1 \\
p_2 \cos x_2 + q_2 \sin x_1 &= r_2
\end{align}$$
so that the equations become interchangeable with a simple index swap $1\leftrightarrow 2$, and also that the cosines (and sines) have matching coefficients. (I'm using $p$, $q$, $r$ to help avoid confusion with the original form of the equations.)
Solving the equations for $\sin x_2$ and $\cos x_2$, then substituting into $\cos^2x_2+\sin^2x_2=1$, yields a polynomial in $\sin x_1$ and $\cos x_1$. Squaring appropriately puts all trig functions to an even power so that we can rewrite sines as cosines to get this quartic polynomial in $k_1:=\cos x_1$:
$$\begin{align}
0 &= 
\left(
  p_2^2 \left(q_1^2 - r_1^2\right) - q_1^2 (q_2+r_2)^2 \right) \left(
  p_2^2 \left(q_1^2 - r_1^2\right) - q_1^2 (q_2-r_2)^2 \right) \\[4pt]
&+4 k_1 p_1 p_2^2 r_1 \left( q_1^2\left(p_2^2 - q_2^2\right) - p_2^2 r_1^2 - 
   q_1^2 r_2^2 \right) \\[4pt]
&-2 k_1^2 \left( q_1^2 \left(p_2^2 - q_2^2\right)\left(p_1^2 p_2^2 - q_1^2 q_2^2\right) 
- p_2^2 r_1^2 \left( 3 p_1^2 p_2^2 - q_1^2 q_2^2 \right) 
- q_1^2 r_2^2 \left(   p_1^2 p_2^2 + q_1^2 q_2^2 \right) \right) \\[4pt]
&-4 k_1^3 p_1 p_2^2 r_1 \left(p_1^2 p_2^2 - q_1^2 q_2^2\right) \\[4pt]
&+\phantom{4}k_1^4 \left(p_1^2 p_2^2 - q_1^2 q_2^2\right)^2
\end{align}$$
The index swap $1\leftrightarrow 2$ gives the corresponding polynomial for $k_2 := \cos x_2$.
From here, one could theoretically invoke the quartic formula to find the possible values of $k_1$ (and $k_2$). Treating the coefficients symbolically creates quite a sprawling mess, so I'll just leave things here.
