Prove that $f(x)=2x-L\cos(Lx)$ has only one root in range $x\in [-1, 1]$ when $L \in [0, \pi/2]$

Question

Prove that $f(x)=2x-L\cos(Lx)$ has only one root in range $x\in [-1, 1]$ when $L \in [0, \pi/2]$

It's simple to show that it has at least one root using Intermediate value theorem, but I find it difficult to show that it has only one root.

Answer 1

For $x<0$ both terms in $f$ are negative, so that there can be no root in $[-1,0]$. If $x>\frac L2$, the first term dominates the second, as $\cos u\le 1$. Thus the roots can only occur inside $(0,\frac\pi4]$. On this interval $f'(x)=2+L^2\sin(Lx)$ is positive, so that $f$ is monotonous on this interval and can only have one root there and thus in the whole of $[-1,1]$.

Answer 2

Hint (but not a solution):

1. try drawing the graph (Desmos is a good tool; include a slider for the value $L$)

2. For any particular $L$, split into two cases: $x < 0$ and $x \ge 0$. On one of these intervals, $f$ is monotone, so the main question is why there are no double-roots in the other interval.