Prove or contradict: Between each two solutions of $\arctan x = \sin x$ exists a solution for $1-\cos x = x^2 \cos x$

Question

Prove or contradict: Between each two solutions of $\arctan x = \sin x$ exists a solution for $1-\cos x = x^2 \cos x$

I have this question in a sample exam and I don't even know what would be a good way to approach this. I though about finding the ranges where the two difference functions have different slopes or something, but I'm not quite sure..

Answer 1

To prove it apply the Standard version of Rolle's theorem for $f\left( x \right)=\arctan \left( x \right)-\sin \left( x \right)$. Link

Answer 2

we have ;

$1-\cos x = x^2 \cos x \implies \cos(x) = \frac1{x^2+1}$

integrate both sides ,

$\int\cos(x) \,dx = \int\frac1{x^2+1}\,dx$

$\sin(x) = \arctan(x) $ $\quad $

Note : i'm ignoring the constant ,because since $0$ is a solution the constants are equal and can be cancelled.

now $g(x) = \arctan(x)-\sin(x)=0$

Apply Rolles theorem,

since at roots the values are equal ie $0$, rolles theorem is applicable and proves that between each zeros of $\arctan(x)=\sin(x)$ there exists a root of $1-\cos(x) =x^2\cos(x)$