This is something we all do in high school but I forgot how to solve such a problem. It recently came up in my theoretical Physics research. I want to find a constraint on the variable $A$ such that the equation $A \cos{x} = x$ has exactly two solutions.
I started with plotting $y=\cos(x)$ and $y= \displaystyle{\frac{x}{A}}$ and so the limiting case of the solution seems to be when the line is tangent to the $cos(x)$ at one point and intersects it once. See Figure. Now, I don't know how to find the required condition. Much appreciate your help. Thanks.
Edited after corrections were pointed out. Added picture.
May be easier is to consider the extrema of function $$f(x)=\frac {\cos(x)}x -A$$ for which $$f'(x)=-\frac 1{x^2}(x \sin (x)+\cos (x))$$
So, we need to find the zeros of function $$g(x)=x \sin (x)+\cos (x)$$ Because ot the $x$ term, $g(x)$ will cancel closer and closer to $n\pi$.
So, make a infinte series expansion $$g(x)=\sum_{k=0}^\infty (-1)^n \frac{\pi n \sin \left(\frac{\pi k}{2}\right)-(k-1) \cos \left(\frac{\pi k}{2}\right)}{k!}\,(x-n\pi)^k$$ Truncate to some order and use power series reversion to obtain for the $n^{\text{th}}$ root using $q=n \pi$ $$\color{blue}{x_{(n)}=q-\frac 1 q\left(1+\frac{2}{3 q^2}+\frac{13}{15 q^4}+\frac{146}{105 q^6}+O\left(\frac{1}{q^8}\right) \right)}$$ Notice that using Morse and Feshbach generalization we could have as many terms as required.
For example, using the above truncated series, converted to decimals, the first solution is $x_{(1)}=2.79849$ while the "exact" solution as given by Newton mathod is $2.79839$ and the result will be more and more accurate.
This gives $$A_{(n)}=\frac 1 q \left(1+\frac{1}{2 q^2}+\frac{13}{24 q^4}+\frac{61}{80 q^6}+O\left(\frac{1}{q^8}\right)\right)$$