Show that the equation $ \cos(x) - kx = 0$ has a unique solution in $[0, \pi/2]$ for all $k>0$

Question

Show that the equation $$f(x;k) \equiv \cos(x) - kx = 0$$ has a unique solution in $[0, \pi/2]$ for all $k>0$.

Answer 1

The function $g(x)=\cos x-kx$ is the sum of two strictly decreasing functions in $[0,\pi/2]$, so it can not have more than one zero.

On the other hand, $g(0)=1>0$ and $g(\pi/2)=-k\pi/2<0$. By Bolzano's theorem, it must have one zero.

Answer 2

Let $f(x) = \cos(x) - kx $ with $k > 0$.

$f(0) = 1$ and $f(\pi/2) =-k\pi/2 < 0 $, so $f$ has at least one root there.

$f'(x) =-\sin(x)-k < 0 $ so $f$ is decreasing, so it can have at most one root.

Therefore $f$ has exactly one root.

Note that the same proof shows the same thing for $f(x) =a\cos(x)-kx $ with both $a$ and $k$ positive.