Why do some trig equations give solutions in terms of roots? In some optimization problems, I started getting very odd answers, and have traced it back to solutions of the following equation:
$$\cos(t)+\tan(t)\sec(t)=0$$
Mathematica tells me that the solution of this equation (for reals between $-\frac\pi2$ and $\frac\pi2$) is $2 \arctan(a)$ where $a$ a root of the equation $-1 - 2 x + 3 x^2 - 4 x^3 - 3 x^4 - 2 x^5 + x^6$. I'm not super interested in the derivation of this specific solution, but more broadly, how is it that trigonometric equations have solutions in terms of these giant unwieldy polynomials? 
Is there a process for deriving said polynomials?
 A: Write your equation in the form
$$\cos(t)+\frac{\sin(t)}{\cos^2(t)}=0$$ and then make the substitution:
$$\sin(t)=\frac{2t}{1+t^2}$$
$$\cos(t)=\frac{1-t^2}{1+t^2}$$
Then you have to solve $${\frac {-{t}^{2}+1}{{t}^{2}+1}}+2\,{\frac {t}{-{t}^{2}+1}}=0$$ for $t$
A: Any of the six trigonometric functions can be written as a rational function of $\sin{x}$ and $\cos{x}$. If we have a rational function of $\sin{x}$ and $\cos{x}$, it can be written as a ratio of polynomials in $\sin{x}$ and $\cos{x}$. So to solve $f(\sin{x},\cos{x})=0$ for a rational function, it suffices to solve $p(\sin{x},\cos{x})=0$ for the particular polynomial $p$ in the numerator of $f$ (in lowest terms, etc.)
But by the tangent half-angle formulae, if $t=\tan{(x/2)}$, we have
$$ \sin{x} = \frac{2t}{1+t^2} , \quad \cos{x} = \frac{1-t^2}{1+t^2}, $$
so the solutions are the solutions of
$$ p\left( \frac{2t}{1+t^2} , \frac{1-t^2}{1+t^2} \right) = 0 . $$
Multiplying this by a power of $1+t^2$ then finally gives a polynomial in $t$.
For example, the equation you give is
$$ 0 = \cos{t} + \tan{t}\sec{t} =  \frac{\cos^2{t}+\sin{t}}{\cos{t}} , $$
so we want to solve
$$ 0 = \cos^2{t} + \sin{t} . $$
Using the tangent half-angle formulae,
$$ 0 = \frac{(1-t^2)^2}{(1+t^2)^2} + \frac{2t}{1+t^2} = \frac{(1-t^2)^2 + 2t(1+t^2)}{(1+t^2)^2} , $$
and the numerator should simplify into the polynomial you've given.
It's simpler in this case to rewrite $\cos^2{t} = 1-\sin^2{t}$, and then you find that you need to solve
$$ 1 + \sin{t} - \sin^2{t} = 0 , $$
which is much easier and won't introduce spurious roots.
A: Short answer is that trig angular relations are themselves ratios of right triangle length of sides ... like $x = f(\theta)= f(t) $ say.. in the given example.
So if $$ x=\tan \theta= \tan t$$
then
$$\cos 2 x = \frac{1-t^2}{1+t^2}$$
etc. and for similar Weierstrass trig relations will define   polynomials in a functional polynomial relationship by eliminating of $\theta$ or $t$, leaving behind polynomials.
To be clear the matter of next obtaining roots of the resulting  polynomial is entirely a  consequential process.
