# How many solutions does $\cos(97x)=x$ have?

How many solutions does $\cos(97x)=x$ have?

I have plot the function. However I don't know how to solve the problem without computer.

Can anyone give a fast solution without a computer?

• It's a GRE Problem. Oct 19, 2018 at 6:08

## 2 Answers

Let us concentrate on $[0,1]$ as suggested by Siminore. The period of $\cos(97x)$ is $\frac{2\pi}{97}$. On $[0,1]$, the function repeats itself $$\frac{1}{\frac{2\pi}{97}}\approx 15.43$$ On each period, the functions $\cos(97x),x$ meet twice, so you get at least $30$ meetings. Since it does almost half a period after its $15^{th}$ before going over $1$, you can convince yourself that on the $0.43$ part, they will meet again, so we have $31$ meetings.

On $[-1,0]$ the function will also repeat approximatively $15.43$ times, however, this time, it won't be enough to get another meeting, so there will only be $30$ on that part. There is then a total of $61$ solutions.

For clarity, they meet $30$ times in $[0,15\times\frac{2\pi}{97}]$ and since $15\times\frac{2\pi}{97}\approx 0.97<1$ and $\cos(97\times 1)\approx -0.925$, the functions will meet once more in $[15\times\frac{2\pi}{97},1]$.

On the other hand, after the $15^{th}$ period of $\cos(97x)$ (Imagine it starts from $0$ and goes backward to $-1$), $x$ is about $-0.97$ and decreasing, while $cos(97x)$ is at $1$ decreasing up to $\cos(-97)\approx -0.925$, not enough for another meeting

• But this is not symmetric, and there is actually one less solution on the $x<0$ side. Nov 3, 2012 at 16:11
• @Jonathan you are right, I should have plotted both side Nov 3, 2012 at 16:14

Since $-1 \leq \cos (97x) \leq 1$, you can restrict your attention to the interval $[-1,1]$, and, even better, to $[0,1]$ by evenness.

If you sketch the graph of $x \mapsto \cos (97x)$, you'll understand that you need to count the "bumps" of this function lying in the upper half-plane, and add 1 since the first "bump" is crossed by the $y$-axis. See the graph here.