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!
Best Answer
[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.