Solving for (or approximating) $x$ in $\frac{x-\sin x}{x^{0.4}} = c$ I'm an engineer and I'm deducting a formula, and I found a BIG problem: I can't isolate $x$.
The formula is:
$$\frac{x-\sin x}{x^{0.4}} = c$$
where $c$ is a constant.
How to find the value for $x$, or at least a good approximation?
 A: The leading term of $x-\sin x$ is $\frac{x^3}6$. This would result in a first approximation of $x=(6c)^{1/2.6}$ for smallish values of $c$.
For larger values of $c$ one would use that $|\sin x|\le  1$ to get approximately $x=c^{1/0.6}$.
A: I would use Newton-Raphson, with $c^{\frac{5}{3}}$ as initial guess. If $c$ is close to $0$ (e.g. in $(-1, 1)$), you could use $\sqrt[2.6]{6c}$ instead.
The function oscillates around $x^{\frac{3}{5}}$ and is nice and smooth. This should work fine.
I tested it for $c= 10$ and within 3 iterations the error was tiny. Depending on the required accuracy, you can easily add more iterations.
A: There is no analytical function that will give you x = f(c). I can recommend Wolfram Alpha to get a quick insight into the problem: (https://www.wolframalpha.com/input/?i=%28x+-+sinx%29+%2Fx%5E0.4+%3D+10+solve+for+x).
You get immediately a plot of your function. Hope that helps a bit
A: Here is the graph of the inverse of
$$f(x)=\frac{x-\sin x}{x^{0.4}}$$
as you can see it is not a function. For some value of $c$ there are more than one $x$, even three in some intervals.
For $c=3$ we have $x_1\approx 4.5;\;x_2\approx 6.35;\;x_3\approx 7.83$.
For $c=3.3$ we have $x\approx 8.55$.
I used GeoGebra, plotted $f(x)$ and to get the inverse I took its symmetric with respect to the bisector of the first quadrant. This gives fine approximations for any $c$.
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