Why does the function $\sin(\sin(x) \cdot x)$ not represent FM? I was trying to figure out a way to represent frequency modulation (FM) with sinusoids, and thus graphed the function $f(x) = \sin(\sin(x) \cdot x)$. Needless to say, the graph was not what I wanted. I couldn't understand why the graph is shaped as such. Since $\sin(x)$ oscillates between $-1$ and $1$, I would've thought that the wavelength of $f(x)$ would not go below $2\pi$. Is there something I'm missing? I could always confirm it mathematically using derivatives and such, but I'm looking for intuition.
P.S. I would also appreciate it if anyone could also share and explain the equation representing FM with sinusoids, but I'll probably ask another question just for that.

 A: The zeroes of $\sin(x\sin x)$ occur at the solutions of $x\sin(x)=\pi k$:

this gives that $\sin(x\sin(x))$ is not periodic (the density of zeroes increases as we move away from the origin)
and explains its non-negativity over a large neighbourhood of the origin.
A: Answering here so I will understand later on when I revisit this topic:
Imagine $f(x)=x\sin(x)$ running through the values from $-x$ to $x$ in one period (as $\sin(x)$ traverses through $-1$ and $1$, but here $x$ also acts as a variable amplitude). It will traverse many points where $x\sin(x)=k\pi$. As $x$ increases down the $x$-axis, the "faster" $f(x)$ traverses the $x$-values as it takes the same amount of time to run through all the values of $x$ (sharper gradient).
When plugging $f(x)$ into another sine function, i.e. $\sin(x\sin x)$, it could be thought as traversing through multiple wavelengths faster and faster as $x$ increases. This can be proven knowing that every time $f(x) = k\pi$, $\sin(f(x)) = 0$. As $x$ increases, the number of points where $f(x) = k\pi$ increases as well, so the frequency of $\sin(x\sin x)$ must increase with $x$.
To model FM, we need to instead set a variable that will represent the change in frequency and vary it in accordance to the signal wave. This takes away the modulated frequency's direct dependence on the $x$-values and solves the problem.
