Quadratic With Periodic Coefficents I've come across a problem that results in the equation:
$t^2 -2t\sin t -2\cos t -2 = 0$
I've tried to do this analytically but I can't figure it out. At this point, I just want to know if it's even possible for something like this to be solved exactly. So can an equation $ax^2+bx+c=0$, where $a, b, \text{or } c$ are a trig function, be solved?
 A: Using graphics or inspection, the zero of function
$$f(t)=t^2 -2t\sin( t) -2\cos (t) -2 $$is just above to $\frac {2\pi}3$
$$f\left(\frac{2 \pi }{3}\right)=-1-\frac{2 \pi }{\sqrt{3}}+\frac{4 \pi ^2}{9}=-0.241108\cdots$$
Expanded as series
$$f(t)=\left(-1-\frac{2 \pi }{\sqrt{3}}+\frac{4 \pi ^2}{9}\right)+2\pi \left(t-\frac{2 \pi }{3}\right)+\left(\frac{3}{2}+\frac{\pi }{\sqrt{3}}\right)\left(t-\frac{2 \pi }{3}\right)^2 +$$
$$\frac 13 \sum_{n=2}^\infty \frac { \left(2 \pi -3 \sqrt{3} (n-1)\right) \sin \left(\frac{\pi  n}{2}\right)-\left(3 (n-1)+2 \sqrt{3} \pi \right) \cos \left(\frac{\pi  n}{2}\right)} { n!} \left(t-\frac{2 \pi }{3}\right)^n$$
Truncate to some low order and use series reversion to obtain, as an approximation,
$$t=\frac{2 \pi }{3}+x-\left(\frac{1}{2 \sqrt{3}}+\frac{3}{4 \pi }\right)
   x^2+\left(\frac{2}{9}+\frac{9}{8 \pi ^2}+\frac{1}{\sqrt{3} \pi }\right)
   x^3-$$ $$\frac{\left(3645+1350 \sqrt{3} \pi +1152 \pi ^2+176 \sqrt{3} \pi ^3\right)
   }{1728 \pi ^3}x^4+O\left(x^{5}\right)$$ with $x=\frac{1}{\sqrt{3}}+\frac{1}{2 \pi }-\frac{2 \pi }{9}$.
This truncated series gives an explicit expression : its decimal representation is
$t=\color{red}{2.132020}087\cdots$ while, as given by Newton method, the solution is $t=\color{red}{2.132020146}\cdots$
