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.

  • $\begingroup$ How can I correct the error in order to tackle the problem? By the way, I don't see why $x \mapsto \sin(\alpha x) - \beta x$ is not differentiable over $\mathbb{R}$? $\endgroup$ – Raito Mar 31 '18 at 18:41
  • $\begingroup$ Alright, nevermind, I see why the implicit function is not differentiable, indeed. $\endgroup$ – Raito Mar 31 '18 at 18:46

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