I am trying to design a function f defined, for every $x>0$, by $ax$ where $a$ is a constant value that I am searching for. Now, the problem is, for a fixed $c>0$, find $a$ such that the equation
$$ax - 2*\sin^2(x/c)=0$$
has only two roots. In other words, find the straight line that is tangent to $2*\sin^2(x/c)$ and passes through 0.
I have been trying to define $a=4\sin(x/c)\cos(x/c)/c$, i.e., the derivative of $2*\sin^2(x/c)$, and reformulate the above problem as finding $c$ such that
$$x4\sin(x/c)\cos(x/c)/c - 2*\sin^2(x/c)=0$$
has two roots. But this does not seem to be simpler. Does someone has any idea?
Your problem is to find a point $P\equiv (x,f(x))$ such that: $$ f'(x)=\dfrac{f(x)}{x}=a $$ with $f(x)= 2\sin^2\left(\dfrac{x}{c}\right)$. since $f'(x)= \dfrac{2}{c}\sin\left(\dfrac{2x}{c}\right)$ you have the equation: $$ x\dfrac{2}{c}\sin\left(\dfrac{2x}{c}\right)=2\sin^2\left(\dfrac{x}{c}\right) $$ that, I fear, we can not solve with finite formula (and has, obviously, infinite roots).