Answer
See the explanation below.
Work Step by Step
Rewrite the equation as $f(x)=0$
$\cos x-x=0, \qquad f(x)=\cos x-x$
$f$ is a difference of continuous functions, so it is also continuous, by Th.8.
Note that:
$ f(0)=1-0=1\quad$which is positive.
$f(\displaystyle \frac{\pi}{2})=-\frac{\pi}{2} \quad$which is negative.
$y_{0}=0$ is a value between $f(a)=f(0)=1$ and $f(b)=f(\displaystyle \frac{\pi}{2})=-\frac{\pi}{2}.$
The Intermediate Value theorem guarantees that there exists a $c\displaystyle \in(0,\frac{\pi}{2})$ for which $f(c)=0$
This means that that $c$ is a solution to the given equation.
(The equation has a solution.)
