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?
On
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 $t=\color{red}{2.132020146}\cdots$
You can reduce the equation to a complicated power function equation by:
$$t^2-2t\sin(t)-2\cos(t)=2$$
Use the complex definitions of the function:
$$t^2-it e^{-i t}+ i t e^{i t}- e^{- i t}-e^{it}=2$$
A transformation $x=it$:
$$-x^2-xe^{-x} -e^{-x} +xe^x-e^x= -x^2-e^{-x}(x+1) +e^x(x-1) =2$$
Now use the limit definition of $e^x$:
$$\lim_{a\to\infty}\left(\frac{a+ x}a\right)^{\pm a}=e^{\pm x}\implies -x^2-\left(\frac{a+ x}a\right) ^{-a}(x+1)+ \left(\frac{a+ x}a\right) ^a(x-1)=2 $$
Therefore your equation can be a power function as $a\to\infty$:
$$\boxed{-x^2-a^a(x+a)^{-a}(x+1)+ a^{-a}(x+a) ^a(x-1)=2}$$
For $a=10$, the expanded polynomial of $x$ is: $$-400000000000000000000 - 400000000000000000000 x - 180000000000000000000 x^2 - 48000000000000000000 x^3 + 16350000000000000000 x^4 + 20442000000000000000 x^5 + 9486000000000000000 x^6 + 2846400000000000000 x^7 + 628140000000000000 x^8 + 107972000000000000 x^9 + 14903420000000000 x^10 + 1678600000000000 x^{11}+ 155353000000000 x^{12} + 11821800000000 x^{13} + 736440000000 x^{14} + 37209600000 x^{15} + 1501950000 x^{16} + 47310000 x^{17} + 1121000 x^{18} + 18800 x^{19} + 199 x^{20} + x^{21} =0\implies x=i t=2.0374…,-1.78203…$$
using $a=90$ gives this result
$$t≈\pm 2.13207…-0.015…i$$
while the true real root of your equation is: $$t=\pm 2.1320…$$
If the inverse of this power function equation can be found, then we will have an inverse function for your problem. For $a\in\Bbb N$, you will have a better polynomial approximation. Please correct me and give me feedback!
Someone more familiar with the Lagrange inversion theorem could help since it can find the series expansion, with closed form coefficients of $x^n-c x$, so maybe the linked theorem also help find the inverse of the boxed equation.