Solve for $x$ where $A = \sin (x) /B + \cos (x)/C$ where $A, B,$ and $C$ are known It's been a few years since I've done trig and I've reduced an equation down to the following:

$$A = \sin (x)/B + \cos (x)/C$$

Is it possible to solve for $x?$

Edit: Diagram and clarification added.

The problem I'm trying to solve is the angle of rotation for a scaled down rectangle inside another rectangle such that all 4 corners touch the original rectangle.



PS: Thank you all for the amazing responses!  This was clearly way above my head.  I'm super jealous and wish I had continued to pursue math after graduation.  I'll definitely have to start brushing up...
This is a linear equation in sine and cosine. Rewrite it as
$$
C\sin x+B\cos x=ABC
$$
and then divide both sides by $\sqrt{B^2+C^2}$:
$$
\frac{C}{\sqrt{B^2+C^2}}\sin x+\frac{B}{\sqrt{B^2+C^2}}\cos x=\frac{ABC}{\sqrt{B^2+C^2}}
$$
Set $s=ABC/\sqrt{B^2+C^2}$ and find an angle $\alpha$ such that
$$
\sin\alpha=\frac{C}{\sqrt{B^2+C^2}},
\qquad
\cos\alpha=\frac{B}{\sqrt{B^2+C^2}}
$$
which is surely possible. Then the equation becomes
$$
\cos x\cos\alpha+\sin x\sin\alpha=s
$$
or
$$
\cos(x-\alpha)=s
$$
that is elementary.

Of course, there's no solution if $|s|&gt;1$.

Note about “surely possible”: the point with coordinates
$$
\left(\frac{B}{\sqrt{B^2+C^2}},\frac{C}{\sqrt{B^2+C^2}}\right)
$$
lies on the unit circle. Connecting it with the origin defines the required angle $\alpha$, considering as the other angle side the positive $x$-semiaxis.
$$
\cos(x - \phi) = \cos x\cos\phi + \sin x\sin \phi
$$
let
$$
\frac{1}{B} = \sin \phi\\
\frac{1}{C} = \cos \phi\\
$$
where $\vert\frac{1}{B}\vert \leq 1 $ and similarly for $\vert\frac{1}{C}\vert \leq 1 $.
thus
$$
\tan\phi = \frac{C}{B}
$$
provided that $C/B$ such that $|\tan \phi| &lt;\infty$ 
then we have
$$
\cos(x-\phi) = A \text{for $\vert A\vert &lt; 1$}
$$
solve for $x$ as usual.