Why is the solution to this sum of two sines so complex? While working on a signal processing problem, I found I needed to solve a certain equation involving a sum of a sequence of sines. Since I wasn't really sure how to do this cleanly, I decided to ask WolframAlpha to solve a simplified version (the first two terms) and see if there was a general pattern I could extrapolate.
To my utter surprise, the equation
$$ \sin(a-x) + 2\sin(b-2x) = 0$$
produced this monstrosity (note that this is just one of two solutions!):



What exactly is going on here? Why is the solution to this equation so absurdly complex?
I ended up solving my problem another way, but I still would love to know what is with this particular equation.
 A: (New result added at the end)
In
$\sin(a-x) + 2\sin(b-2x) = 0
$,
let $y = a-x$
so
$x = a-y$.
This becomes
$\begin{array}\\
0
&=\sin(y)+2\sin(b-2(a-y))\\
&=\sin(y)+2\sin(b-2a+2y)\\
&=\sin(y)+2\sin(c+2y)
\qquad c = b-2a\\
&=\sin(y)+2(\sin(c)\cos(2y)+\cos(c)\sin(2y))\\
&=\sin(y)+2(\sin(c)(1-2\sin^2(y))+2\cos(c)\cos(y)\sin(y))\\
&=\sin(y)+2u(1-2\sin^2(y))+4v\cos(y)\sin(y))
\qquad u = \sin(c), v = \cos(c)\\
\end{array}
$
so
$4v\cos(y)\sin(y)
= -\sin(y)-2u(1-2\sin^2(y))
$
or,
if $s = \sin(y)$,
$4vs\sqrt{1-s^2}
= -s-2u(1-2s^2)
= -s-2u+4us^2
$.
If we square this,
the result is a quartic
$4v^2s^2(1-s^2)
=(-s-2u+4us^2)^2
$
or
$4(1-u^2)s^2(1-s^2)
=(-s-2u+4us^2)^2
$.
This is a quartic in $s$
which will probably
have a very messy solution
like the one you have.
(added later)
If $c = \pi/4$
so $u^2 = \frac12$,
the resulting equation
according to Wolfy is
$-10 s^4 + 4 \sqrt{2} s^3 + 9 s^2 - 2 \sqrt{2} s - 2 = 0
$
which, surprisingly,
has roots
$\pm \dfrac{\sqrt{2}}{2}
\approx \pm 0.70711$
and
$\dfrac15(\sqrt{2}\pm2\sqrt{3})
\approx -0.40998, 0.97566
$.
