Existence of two solutions I am having a problem with the following exercise.
I need to show the $x^2 = \cos x $ has two solutions.
Thank you in advance.

Assuming you want to show that $f(x) = \cos x - x^2$ has two solutions over $\mathbb R$, you could try the following approach: Since both $\cos x$ and $x^2$ are even functions, if $f(x)$ has a solution $x > 0$, $-x$ is also a solution. Therefore, it is sufficient to prove that $f(0) \neq 0$ and there is exactly one solution $x \in ]0,\infty[$.
The first part is easy, since $f(0) = \cos 0 - 0^2 = 1$. So, split up the interval $]0,\infty[$ into two parts: $]0,1]$ and $]1,\infty[$. In the following, we will use three well-known properties of $\cos x$, namely:


*

*$\cos x$ is strictly decreasing on $[0,\pi]$.

*$|\cos x| \le 1$ for all $x \in \mathbb R$.

*$\cos$ is continuous.


In the first interval, we have that $1 = \cos 0 > \cos 1$. On the other hand, $0^2 = 0 < 1$ and $1^2 = 1 > \cos 1$, so $f(0) = 1$ and $f(1) < 0$. By continuity of $\cos$ and $x^2$, $f$ is continuous. Hence by the Intermediate Value Theorem, the equation $f(x) = 0$ has a solution in $[0,1]$, and that solution is not $0$.
Furthermore, let $x > 1$. Then $f(x) \le 1- x^2 < 0$ by property (2), so there are no solutions for $x > 1$ (and, by $f(x)$ even, $x < -1$).
It remains to show that there is no further solution in $[0,1]$. But this follows from the fact that $f(x)$ is strictly decreasing on $[0,1]$.
