Solving $\sin(f(x))=g\left(x\right)$ for $x$

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This may be a very dumb question but whenever I try to look it up all my results are about derivatives and integrals. Is there any general way of just solving the equation $\sin(f(x))=g(x)$ for $x$? Currently I'm trying to solve

$$\frac{\sqrt{x-9}\sqrt{x-1}-3}{x}=\sin\left(\frac{3\sqrt{x-9}+\sqrt{x-1}}{2x}\right)$$

and don't know even the first step. Let's just define $f(x) = \frac{\sqrt{x-9}\sqrt{x-1}-3}{x}$ and $g(x) = \frac{3\sqrt{x-9}+\sqrt{x-1}}{2x}$ so I don't have to keep writing everything out. The only method I can think of is to do the following:

$$f(x) = \sin(g(x))$$

$$\arcsin(f(x)) = g(x)$$

$$\int \frac{f'(x)}{\sqrt{1-f(x)^2}} \, dx = g(x)$$

And just pray that there is a second way to integrate the left and end up with a solvable equation. Is there are good way to do this that I'm missing? Sorry again if this is a dumb question.