Rolle's theorem problem prove by contradiction

Given $0≤b$ we need to prove the $cos(x)=bx$ has one and only solution in $[0,0.5π]$
So I first checked for $b=0$ and we get

$cos x-bx=0$

$cos x=0$

and then we get $x=0.5π$ which is our only cut point

then check for $b>0$

$f(x)=cos x-bx$

$f(0)=cos 0-0=1$

$f(0.5π)=cos 0.5π-0.5πx=$$frac{-bπ}{2}$

which according to the intermediate value theorem it has to have at least one solution (negative and positive)

now after doing the derivative we get $f′(x)=-sin x-b$ is always negative so it means the slope is negative and it keeps going down and never up so it has exactly one solution.

but what I am practicing is trying to prove by contradiction, so what I tried is

suppose we are going to prove by contradiction then we can assume that $f(x)$ has 2 solutions , and then according to Rolle’s theorem the derivative should have 1 solution so we get $f′(x)=-sin x-b$ and then we put $f′(x)=0$ to find the point and we get $b=-sin x$ I checked on some graphs on the internet and saw that it really does not cut the $x$-axis(in $[0,0.5π]$) as long as $b>0$ but I could not find a way to prove this not mathematical way and not by explaining.

Appreciate any help on how to prove this by contradiction or if my approach to it by contradiction was right.

And I am sorry for my English mistakes – hope it is understandable.