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.
(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 $.