Let be $(\alpha, \beta) \in \mathbb{R}^2$.
We have $(E) : \lvert \sin (\alpha x) \rvert = \beta x$ an equation where $x \in \mathbb{R}$ is the unknown.
If we consider $\mathcal{S}$ the set of solutions of $(E)$, I would like to compute the cardinal of $\mathcal{S}$, eventually get a closed form of all solutions.
Here is my approach:
Necessarily, if such $x \in \mathbb{R}$ exists, then $\beta x \in [0, 1]$.
Let us suppose $\beta x \in [0, 1]$, otherwise, there'll be no solution.
Let be $x \in \mathbb{R}$ such that $\lvert \sin(\alpha x)\rvert = \beta x$.
Let us suppose $\sin(\alpha x) > 0$, without loss of generality.
Now, $\sin(\alpha x) = \beta x$ can be differentiated in respect for $x$:
$\begin{align*} & \alpha \cos(\alpha x) = \beta \\ \text{i.e. } & \cos(\alpha x) = \dfrac{\beta}{\alpha} \end{align*}$
Thus, we have $x = \dfrac{1}{\alpha} \arccos \dfrac{\beta}{\alpha} + 2n\pi, n \in \mathbb{Z} \quad (1)$ or $x = -\dfrac{1}{\alpha} \arcsin \dfrac{\beta}{\alpha} + 2n\pi, n \in \mathbb{Z} \quad (2)$.
By symmetry, if $\sin(\alpha x) < 0$, we also have: $x = \dfrac{1}{\alpha} \arccos -\dfrac{\beta}{\alpha} + 2n\pi, n \in \mathbb{Z} \quad (3)$ or $x = -\dfrac{1}{\alpha} \arcsin -\dfrac{\beta}{\alpha} + 2n\pi, n \in \mathbb{Z} \quad (4)$.
Now, let us verify that those are indeed solutions of $(E)$.
For $(1)$:
$\begin{align*} \lvert \sin \alpha x \rvert & = \Big \lvert \sin \arccos \dfrac{\beta}{\alpha} \Big \rvert \\ & = \Big \lvert \sqrt{1 - \left(\frac{\beta}{\alpha}\right)^2} \Big \rvert \\ & = \dfrac{\sqrt{\alpha^2 - \beta^2}}{\alpha} \end{align*}$
But, at the same time, I don't see very well how $\dfrac{\beta}{\alpha} \arccos \dfrac{\beta}{\alpha} + 2n\beta\pi, n \in \mathbb{Z}$ could be equal to some square root.
Your error is assuming that $\sin(\alpha x) = \beta x$ can be differentiated.
This is true at a specific value of $x$, not a range of values.