Simultaneous equation of trig functions If I have the two following equations:
$$\begin{align}
A &= F\sin(x)+ G\cos(x)\\
B &= G\sin(x)+ F\cos(x)
\end{align}$$
and $A$,$B$,$F$,$G$ and are all constants, what is the easiest method of calculating $x$?
I've had a look at Cramers rule and the $R\cos(x-\alpha)$ rule. I need to find the most effective method for a program I am writing!
 A: Use $A^2+B^2=F^2+G^2+2FG\sin 2x,\,A^2-B^2=(G^2-F^2)\cos 2x$ to get $\sin 2x,\,\cos 2x$.
A: $$
\left\{ \matrix{
  A = F\sin x + G\cos x \hfill \cr 
  B = F\cos x + G\sin x \hfill \cr}  \right. \Rightarrow \left\{ \matrix{
  F\sin x + G\cos x = A \hfill \cr 
  G\sin x + F\cos x = B \hfill \cr}  \right.
$$
is a linear system in $\sin x $ and $\cos x$.
To solve it programmatically you will first solve the linear system in $\sin x, \, \cos x$, and after determine if the the couple of values obtained are "trigonometrically" congruent so that they may correspond to a single value for $x$. That can happen only if $\sin ^2 x+ \cos ^2 x = 1$.
Therefore
a) compute the determinant of the coefficients $D_c = F^2-G^2$
b) if $D_c=0$ then
 - that means $F = \pm G$, choose the sign according to actual values
 - compute the det of one minor of the complete matrix, e.g. $GB-FA = F(\pm B-A)$
 - if that is also null, then the two equations are dependent, and you can solve
  just one of them, e.g. $\sin x = A/F \mp \cos x$ , else there's no solution
 - thereafter, to have solutions in $x$ it must be
$$
\eqalign{
  & 1 = \cos ^2 x + \sin ^2 x = \cos ^2 x + \left( {A/F \mp \cos x} \right)^2 \; \Rightarrow   \cr 
  &  \Rightarrow \;2\cos ^2 x \mp 2A/F\cos x + \left( {A/F} \right)^2  - 1 = 0\; \Rightarrow   \cr 
  &  \Rightarrow \;\cos x = c_1 ,c_2  \cr} 
$$
 - if $ c_1 ,\, c_2 \in \mathbb R $ and one or both are in the range $[-1,1]$ you can solve via $\arccos$
to get from one to four solutions (apart periodicity), else no (real) solution .
c) elif $D_c \ne 0$ then   you have a unique solution, e.g. via Cramer rule,
$$
\left\{ \matrix{
  \sin x = S \hfill \cr 
  \cos x = C \hfill \cr}  \right.
$$
 - again, to have a solution in $x$, it must be
$$
 - 1 \le S,C \le 1\; \cap \;S^2  + C^2  = 1
$$
and solve one of them via $\arccos$ or $\arcsin$, taking into account the respective signs, i.e. via $\arctan$ on the four quadrants.
--- concerning your comment  ---
How to implement your code depends on what you know a priori about the input data 
$A,B,F,G$.
If you are sure that they lead to a unique solution in $x$, because, for example, they come from
a physical system in which $x$ is well defined, then you can use just one equation or a better
suited combination of the two.

My answer concerns the situation in which the input data are general. 

In this case, you shall avoid answers such as "division by 0", or that the program enters
in loop trying to find solutions that cannot be find, or provide you with a solution
 which is not unique, or other inconvenients.
If you want to avoid that, I am sorry but cannot devise any simpler way that the one indicated,
which essentially comes to distinguish two cases $F^2 - G^2=0$ and $F^2 - G^2 \ne 0$.
In the second case, the result from Cramer's formula is not much complicated
$$
\left\{ \matrix{
  \sin (x) = S = {{FA - GB} \over {F^{\,2}  - G^{\,2} }} \hfill \cr 
  \cos (x) = C = {{FB - GA} \over {F^{\,2}  - G^{\,2} }} \hfill \cr}  \right.
$$
and you have to check the three conditions
$ - 1 \le S \le 1$ and $ - 1 \le C \le 1$  and $S^2  + C^2  = 1$
for a real solution in $x$ to exist.
A: If $F\ne G$, then subtracting the second equation from the first gives
\begin{eqnarray}
(F-G)\sin x-(F-G)\cos x&=&A-B\\\
\sin x-\cos x&=&\frac{A-B}{F-G}\\\
\sqrt{2}\left(\frac{1}{\sqrt{2}}\sin x-\frac{1}{\sqrt{2}}\cos x\right)&=&\frac{A-B}{F-G}\\\
\sqrt{2}\left(\sin x\cos\frac{\pi}{4}-\cos x\sin\frac{\pi}{4}\right)&=&\frac{A-B}{F-G}\\\
\sqrt{2}\sin\left(x-\frac{\pi}{4}\right)&=&\frac{A-B}{F-G}\\\
\sin\left(x-\frac{\pi}{4}\right)&=&\frac{\sqrt{2}}{2}\cdot\frac{A-B}{F-G}\tag{1}
\end{eqnarray}
In the case $F=G$ it follows that $A=B$ and the equation reduces to the form
$$\sin x+\cos x=C $$
which can be reformed into
$$ \cos\left(x-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}C $$
ADDENDUM Due to the comment by @GCab
If $F\ne-G$ and the first equation is added to the second a similar process gives the result
$$ \cos\left(x-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\cdot\frac{A+B}{F+G} \tag{2}$$
These equations can be solved for $x$ over whichever range is required.
In the case that $G=-G$ it follows that $A=-B$ and the equation will be of the form
$$\sin x-\cos x=C $$
which has solution 
$$ \sin\left(x-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}C $$
A: Given $A$, $F$, and $G$, you're limited to only one or two valid angles: using the $R$-$\alpha$ method lets you convert it to $x = \alpha\pm\arccos(A/R)$.  You can get a similar answer from $B$, $F$, and $G$.  warning: your computer will probably not have the precision necessary to get the exact same answer both times, even if the answers would, absent rounding errors, be compatible.
A: Each of the two equations determines the value of $x$ for a fitting value of $A$, so if the values are distinct the system is uncompatible. You have 
$$\frac{A}{\sqrt{F^2+G^2}}=\frac{F}{\sqrt{F^2+G^2}}\sin(x)+\frac{G}{\sqrt{F^2+G^2}}\cos(x)$$
$$\frac{A}{\sqrt{F^2+G^2}}=\sin(\alpha+x)$$ where $\cos(\alpha)=\dfrac{F}{\sqrt{F^2+G^2}}$ and $A$ a convenient value.
Similarly you have $$\frac{B}{\sqrt{F^2+G^2}}=\cos(\alpha-x)$$ and for suitable $B$ you do have compatibility or not of your system.
