Show that $\exists x \in (0, \pi/2)$ such that $\cos(x)=x$ Let $f(x) = \cos(x)$, $g(x)=x$, both functions are continuous. $f(0)=1$, $f(\pi/2)=0$, so, by the Intermediate Value Theorem, for any $z\in [0,1]$, there exists $c\in [0,\pi/2]$ such that $f(x)=z$.

This should be simple to prove, but for some reason I have a problem with IVT, don't know why. Would appreciate some help.

I think that the fact that $g(x)=x$ is onto $[0,1]\subset [0, \pi/2]$ should be used here somehow.
Consider $h(x)=\cos(x)-x$. Then, $h(0)=1$ and $h\left( \displaystyle\frac{\pi}{2}\right)=-\displaystyle\frac{\pi}{2}$. Hence, $h(0)&gt;h\left( \displaystyle\frac{\pi}{2}\right)$ and, clearly, $h(x)$ is continuous. For the IVT, there exist a $z\in\left( 0,\displaystyle\frac{\pi}{2}\right)$ such that $h(z)=0$. 

$h(z)=0$ if and only if $\cos(z)-z=0$ if and only if $\cos(z)=z$
The proof is immediate by the use of graphs of $y=\cos x$ and $y=x$ in the given interval. See this Wolfram Alpha link for plots