# 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.)

• Some brute force symbol-crunching with Mathematica gets this to a degree-$4$ polynomial in $\cos x_1$ (and likewise $\cos x_2$). While there is a "quartic formula" of sorts (more of a strategy) to get to a closed form, that form is crazy-complicated. Would there happen to be some relations among the $a_i$, $b_i$, $c_i$ that might help simplify things? (For instance, if $a_1^2b_2^2=a_2^2b_1^2$, then the quartic collapses into a simple(ish) quadratic.)
– Blue
May 18, 2020 at 17:08
• @Blue, generally there is no relation among $a_i, b_i, c_i$. May 19, 2020 at 6:10

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.