Function taking on all values thrice

Starting from the following idea:

$$g(x)=\sin x + \frac{2}{3\pi}x$$

g(x) plot

we can adjust the constant for x in such way that

$$f(x)=\sin x + Kx$$

fullfills the given condition.

The value of K can be easily found imposing that:

$$\begin{cases}(\sin x)'=\cos x=-K\\ Kx=-\sin x\end{cases}$$

$$\implies tanx=x \implies x\approx4.49340945790906 \quad K=-\cos x \approx 0.21723362821123...$$

f(x) plot

I like Jack's description, draw a fixed sine curve and find tangent lines through the origin. In this case, I am finding a tangent point with $2 \pi < x < \frac{5 \pi}{2}.$ Under the circumstances, the slope $K$ comes out positive, with $K \approx 0.128374554, $ solution of $$K \left( 2 \pi + \arccos K \right) = \sqrt {1 - K^2}$$ The $x$ value for the tangent is about $7.725251838,$ just below $\frac{5 \pi}{2} \approx 7.853981635$$

This time, we get each value assumed by the function five times, the function being $\sin x - K x.$

Alright, I widened out to include $\pm 14,$ and clicked so it shows the roots and critical points. It says there are critical points at $x \approx \pm 14.008$