Suppose that $0<t<1$ has been given to us. Prove that $f(x) = \sin(x) – tx$ has a solution in $(0,\pi)$
I proposed this statement when I was trying to answer one of the questions on MSE. I tried to prove it using the Intermediate Value Theorem and Banach's fixed point theorem. It turned out that both approaches became very similar to each other.
I have already proven that this is true. So, I'm just proposing it as a fun calculus/analysis question on MSE because it wasn't asked before and it's a more or less difficult question compared to how innocent it looks.
Best Answer
An equivalent statement is that $\operatorname{sinc}x:=\frac{\sin x}{x}$ (defined as $1$ at $x=0$ by continuity) achieves every value $\in (0,\,1)$ for some $x\in (0,\,\pi)$. It suffices by continuity to note $\operatorname{sinc}0=1,\,\operatorname{sinc}\pi=0$.