How can I invert the function $f(\theta)=\frac{\theta-\sin(\theta)\cos(\theta)}{a-\frac{1}{2}\sin(2a)}$ where $a$ is some constant? Even for some small interval like $0 \le \theta \le \pi/2$ and $0 \lt a \le \pi/2$ ?
Edit: I'm a software engineer, not a mathematician, so I attempted to find the inverse using WolframAlpha, which plots the inverse, but says that there is "no result found in terms of standard mathematical functions".
Let's get rid of the $a - \frac12 \sin(a) \cos(a)$ because it's just a positive constant, and write $g(x) = x - \sin(x) \cos(x) = x - \sin(2x)/2$. Since the derivative $g'(x) = 1 - \cos(2x) \ge 0$ with $g'(x) = 0$ only at a discrete set of points, $g$ is strictly increasing and therefore has an inverse. Since $g(x) \to +\infty$ as $x \to +\infty$ and $g(x) \to -\infty$ as $x \to -\infty$, the inverse is defined on all real numbers. There is no closed form for the inverse, but you can compute it numerically in many ways. For example, near $x=\pi/2$ we have the series $$ g(x) = \frac{\pi}{2} + 2 \left(x - \frac{\pi}{2}\right) - \frac{2}{3} \left(x - \frac{\pi}{2}\right)^3 + \frac{2}{15} \left(x - \frac{\pi}{2}\right)^5 + \ldots $$ and we can use reversion of series to get $$ g^{-1}(y) = \frac{\pi}{2} + \frac{1}{2} \left(y- \frac{\pi}{2}\right) + \frac{1}{24} \left(y- \frac{\pi}{2}\right)^3 + \frac{1}{120} \left(y- \frac{\pi}{2}\right)^5 + \ldots$$ convergent for $y$ near $\pi/2$.
Over the reals, the function $f$ is injective. Its inverse function does exist therefore.
1.)
A simple usual way could be Lagrange inversion.
2.)
For closed forms, we have the following.
$$f(\theta)=\frac{\theta-\sin(\theta)\cos(\theta)}{a-\frac{1}{2}\sin(2a)}$$
Let $y$ a real variable.
$$\frac{\theta-\sin(\theta)\cos(\theta)}{a-\frac{1}{2}\sin(2a)}=y$$
As usual for calculating the inverse function in elementary algebra, we want to determine $\theta$ in dependence of $y$.
$a-\frac{1}{2}\sin(2a)\to c$: $$\frac{\theta-\sin(\theta)\cos(\theta)}{c}=y$$ $$\frac{\theta-\frac{-i}{2}(e^{\theta i}-e^{-\theta i})\cdot\frac{1}{2}(e^{\theta i}-e^{-\theta i})}{c}=y$$ $$\frac{\left(e^{\theta i}\right)^4i}{4c}+\theta\left(e^{\theta i}\right)^2-y\left(e^{\theta i}\right)^2-\frac{i}{4c}=0$$
We see, this is an irreducible polynomial equation involving more than one algebraically independent monomials ($\theta,e^{\theta i}$). We therefore don't know how to solve the equation by rearranging for $\theta$ by applying only finite numbers of only elementary operations (means elementary functions).
$\theta\to\frac{x}{i}$: $$1+4cx\left(e^x\right)^2-4cyi\left(e^x\right)^2-\left(e^x\right)^4=0$$
Now we see, the equation seems to be not in a form to solve it in terms of Lambert W or Generalized Lambert W of Mező/Baricz.
Only for the case that $c\neq 0$ is algebraic:
For algebraic $y$, this is an irreducible algebraic equation involving both $x,e^x$. Assuming Schanuel's conjecture is true, the theorem of Lin 1983 implies that such kind of equations don't have solutions except $0$ that are elementary numbers. This means that the inverse function of the function $f$ cannot be an elmentary function if $c\neq 0$ is algebraic.
3.)
As already said by others, this is a kind of Kepler's equation. See Inverse of $f(x)=\sin(x)+x$ and i.a. my answer there.
Already Liouville wrote that he proved that the inverse function of the function in Kepler's equation isn't an elementary function.
