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.

$f(x) = x^2 - \cos x$ is a continuous function. Since $f(0) = -1$ and $f(\frac{\pi}{2}) = \frac{\pi^2}{4}$, $f$ has at least one zero in the interval $(0, \pi/2)$. Its derivative $f'(x) = 2x + \sin x$ is strictly positive in the interval $(0,\pi/2)$ so $f$ is strictly increasing and we conclude that $f$ has exactly one root $f(x_0) = 0$ in the interval $(0,\pi/2)$.
For $x \geq \pi/2$, $f(x) \geq \frac{\pi^2}{4} - 1 > 0$, so $f$ has no roots in $[\pi/2, +\infty)$.
Since $f$ is symmetric with respect to the $y$-axis, meaning $f(x) = f(-x)$, $f$ has only two roots -- $x_0$ and $-x_0$.
