How to find the number of solutions of $6|\cos x|=x$?

Question

Now, I think the only way to solve this problem for a high school student was graphically.

However using pen and paper to draw the graph, it was virtually impossible to justify or refute the existence of the "Fourth" solution.

Using desmos, I realised that we must must work on the the existence of the Fourth solution analytically because $ y=x$ is really close to $6|\cos x|$ at the potential Fourth solution.

You can see for yourself.

So, this being a high school problem, is there any way to predict whether $y=x$ will intersect $6|\cos x|$ or not?

With enough zoom we can see that $y=x$ does not intersect $y=6|\cos x|$. However, how could have I predicted this with a pen and paper?

I am acquainted with calculus but had no clue how to go about it.

Thanks for your time!

Answer 1

[Using some calculus]

It's enough to consider $3\pi/2 \lt x \lt 2\pi$. In that range the cosine is positive, so we can dispense with the absolute value and consider the function $f(x) = x - 6\cos(x)$, show that its minimum is positive and that its second derivative is positive.

To get the minimum, we set the derivative equal to 0:

$$f'(x) = 1 + 6\sin(x) = 0 \implies x = \sin^{-1}(-1/6) \approx 6.1138 > 6,$$ remembering the restriction on the values of x we are considering. Since the cosine cannot be greater than 1, $f(x) = x - 6\cos(x) > x - 6 > 0$ at $x \approx 6.1138$, so the extremum is positive.

The second derivative is $f''(x) = 6\cos(x) > 0$ since the cosine is positive in the interval we are considering, so the extremum is a minimum and the function is concave up throughout that interval.

Therefore $f(x)$ has a positive minimum and is concave up, so it is never 0 on the given interval.

Answer 2

I know this not necessarily relevant, but some high school students take calc. Keeping this in mind what I would suggest to a student is that this is the same as solving $6|\cos(x)| - x =0$ which itself is equal to two problems for each positive and negative side of the absolute values.

For example the positive $6\cos (x) - x = 0$ then there is a Taylor expansion of the cosine and take it to the forth power for example and solve that polynomial for its roots. I think does not solve the problem with pen and paper but it is an approach that also is valid. (Also you do not have to really solve the polynomial you. just have to say something about it's roots)

Answer 3

Here is my approach for solving it analytically: It's a far stretch from proof, but surely sufficient for high school maths :)

So if, at the fourth solution, $g(x)=6\cdot\vert\text{cos}(x)\vert$ crosses $f(x)=x$, there would either need to be a 5th solution or $g$ is precisely tangent to $f$ at this point, i.e., has slope $g\text{'}(x)=1$.

Case 1: $g(x)=f(x)$ has only four solutions, so we have $g\text{'}(x_4)=f\text{'}(x_4)=1$. We know by heart, that, in the relevant domain: $$g\text{'}(x)=6\cdot(\text{cos}(x))\text{'}(x)=-6\cdot\text{sin}(x)$$

This tells us about the unknown $x_4$ that $$-\text{sin}(x_4)=1\backslash 6\iff x_4=-\text{sin}^{-1}(1\backslash 6)=-0.1674480792+n\cdot2\pi$$ for some $n\in\mathbb{Z}$. So is there some $n\in\mathbb{Z}$ such that $\text{cos}(-0.1674480792+n\cdot2\pi)=-0.1674480792+n\cdot2\pi$? Plugging in a few small integer values for $n$ we quickly see that the answer is no! Proceeding to case 2:

Case 2: There exists not only a 4th, but also a fifth solution $x_5$ to $g(x)=f(x)$, where we see on the intuitively drawn graph that $x_4$ and $x_5$ must be very close to each other. The point that we calculated before, where $g\text{'}(x)=f\text{'}(x)$, i.e., where the tangent of $g$ has the same slope as the tangent of $f$, must in this case lie above $f$ (Draw a picture!). So does there exist some $n\in\mathbb{Z}$ such that $\text{cos}(-0.1674480792+n\cdot2\pi)>-0.1674480792+n\cdot2\pi$? By plugging in small $n$, we see that the answer is no in the relevant domain! (There does exist one solution but it's between the second and third intersection of $g$ and $f$, i.e., doesn't tell us anything about the fourth or postulated fifth one)

So $g(x)=f(x)$ only has three solutions.