For what $k$ does $\sin(x)$ hit $kx$ 9 times I saw this problem online and I am pretty stuck with it, here is the problem

For what $k$ does $kx$ intersect $\sin(x)$ exactly 9 times?

My first thought when looking at this problem was to divide out the $x$. I would then be left with $\sin(x)/x=k$
I then tried to find some clever limit or derivative to get the answer, but I am stuck. I assume there is some clever calculus I can do to get this answer, but I can't find it.
Thanks!
 A: It is obvious that $0$ is a Point of Intersection.
Then Every Positive Point of Intersection will have a Matching Negative Point of Intersection.
Hence we want 4 Positive Points of Intersection.
This Image shows that line with Positive $k$.

The line must be a tangent somewhere near the third Positive Peak.
The line is a tangent to the $y=\sin{x}$ curve , which has Derivative $dy/dx=\cos{x}=k$.
The line Passes through $(0,0)$ , hence the slope is $y/x=k$.
Hence , we get $dy/dx=\cos{x}=k=y/x=\sin{x}/x$.
Equivalently : $\tan{x}=x$

Solving that , the Values are :
X = 0
X ≈ 4.49341
X ≈ 7.72525
X ≈ 10.9041
X ≈ 14.0662
The third Peak neighbourhood criteria gives X ≈ 14.0662.
Here $k=\cos{x}=0.070907387605$

The values of $x$ & $k$ (& $\sin{x}$ & $\cos{x}$) work here !
Moving on :
When we look at Negative values of $k$, The Situation is entirely Different.
The Purple line has 3 Positive Points of Intersection, the Grey line has 5 Positive Points of Intersection.

Between these two Extremities , we have the Black line with Exactly 4 Positive Points of Intersection.
Each Black line uses Negative value of $k$.
To get the range , we must evaluate the Extremities , that is , we check the tangent line near the second & the fourth Negative Peaks.
