Number of real roots of equation

Question

Given $\cos(x)=ax$, express the number of real roots as a function of $a$.

I've found that the sign of $a$ doesn't matter since $cos(x)$ is an even function, and that the number of roots increases as the value of $a$ decreases, for infinitely many roots as $a$ approaches zero. But I have no idea where to go from there.

I know that $\cos(x)=x$ is unsolvable analytically, so is this question also only solvable numerically?

Answer 1

Since $x=0$ is never a solution, the equation $\cos x = a x$ is equivalent to $\frac{\cos x}{x}=a$

Then the number of solution of the equation reduces to finding the maxima and minima of the function $f(x)=\frac{\cos x}{x}$. For example, in the graph we can see that there exists $4$ solutions for $a = 0.1612\ldots$ (the top horizontal), $5$ solutions between the two horizontals, $6$ solutions for $a = 0.1067\ldots$ (the bottom horizontal).

To find those critical points you need to solve $f'(x)=0$, which is equivalent to $\cot x = -x$. This equation has not an analytic solution but the roots can be found numerically.

Answer 2

Not really an answer, but this Geogebra screenshot shows how beautiful this question can become, for values $a \approx 0.35$ and $a \approx 0.15$ (the second one is shown as "b" in the screenshot):