Analysis - Prove that there exists a smallest positive number $p$ such that $\cos(p) = 0$ 
Prove that there exists a smallest positive number $p$ such that $\cos(p)  = 0
$.

I think I'm supposed to use either Rolle's theorem or the Mean Value Theorem but I'm not sure how, any help would be appreciated. Thanks!
 A: $\cos$ is a continous function, the set $\{0\}$ is closed, and therefore $\cos^{-1}(\{0\})$ is a closed set.
We conclude that $\cos^{-1}(\{0\})\cap[0,\infty)$ is closed. This set is equal to $\cos^{-1}(\{0\})\cap (0,\infty)$ (because $\cos(0)\neq 0$).
This set is bounded from below and must therefore have a greatest lower bound $\alpha$. Recall that the greatest lower bound of a set is always an adherent point for the set. Since $\cos^{-1}(\{0\})\cap(0,\infty)$ is closed we conclude $\alpha\in \cos^{-1}(\{0\})\cap(0,\infty)$. Therefore the set $\cos^{-1}(\{0\})\cap (0,\infty)$ has a minimum as desired.
A: I'm not sure the upvoted answer is correct. He/She is implicitly using the assumption that a strictly decreasing sequence of positive real numbers necessarily converges to zero. (i.e. that the strictly decreasing sequence of roots of cos(x) converges to zero). The sequence 
$$
 1, \frac{1}{1.1}, \frac{1}{1.11}, \frac{1}{1.111}, \frac{1}{1.11111}, \dots
$$
is strictly decreasing but is bounded below by $\frac{1}{2}$ so can't converge to zero.
It can be modified as follows. First, since $\cos(0) = 1 > 0$ and $\cos(2) < 0$, the intermediate value theorem tells us that there is a zero of $\cos(x)$ in the interval $(0,2)$. In particular, the set S of positive zeros of the cosine function is nonempty. Also, since we are only considering POSITIVE zeros, this set is bounded below by $0$ and thus has an infimum $t \geq 0$. Choose a sequence $x_n$ of elements of S that converges to $t$. Then by continuity, $$\cos(t) = \cos(\lim x_n) = \lim \cos(x_n) = \lim 0 = 0$$. Thus $\cos(t) = 0$ and since from above $t\geq 0$ and $\cos(0)= 1$ (i.e. not $0$), $\ t > 0$.
A: There are only finitely many values of $0<x<2\pi$ such that $\cos x = 0$
One of them must be the smallest
A: How about you restrict the domain to I=(0,pi) and show that since cos(x) is monotonic on I, it is one to one, and therefore only one value, p=pi/2, would be the solution on I since cos(a)=cos(b)=0 --> a=b, the solution p in I is unique and since any other positive solution "q" would have to be outside the interval I, pi/2=p
