Applying the IVT to prove a root exists

The function $f(x) = x - \cos(x)$ is a difference of the polynomial $x$ and the trigonometric function $\cos(x)$, both of which are continuous everywhere. Since a difference of continuous functions is continuous, $f$ is continuous on the closed interval $[0, \pi/2]$, so the hypotheses of the IVT are satisfied on this interval.

Evaluate $f$ at the endpoints: $f(0) = 0 - \cos(0) = -1$, and $f(\pi/2) = \pi/2 - \cos(\pi/2) = \pi/2 - 0 = \pi/2 \approx 1.57$. Since $f(0) = -1 < 0$ and $f(\pi/2) \approx 1.57 > 0$, the value $N = 0$ lies strictly between $f(0)$ and $f(\pi/2)$.

By the Intermediate Value Theorem, since $f$ is continuous on $[0, \pi/2]$ and $0$ is between $f(0)$ and $f(\pi/2)$, there exists at least one $c$ in $(0, \pi/2)$ such that $f(c) = 0$. This proves that the equation $x - \cos(x) = 0$ has a root in $(0, \pi/2)$.