Show that x^2 = x*sin x + cos x for exactly two real numbers x

Consider the equation

$x^2 = x \sin x + \cos x.$

Show that there are two values of $x \in \mathbb{R}$ such that the equation is satisfied.

Proof. Let $f(x) = x^2 - x \sin x - \cos x$ . (We want then to find the zeros of this function since these will be the points that $x^2 = x \sin + \cos x$ .) Then,

$f'(x) = 2x - \sin x - x \cos x + \sin x = x(2 - \cos x).$

Since $2 - \cos x \neq 0$ for any $x$ (since $\cos x \leq 1$ ), we have

$f'(x) = 0 \quad \iff \quad x = 0.$

Then, $f$ is continuous and differentiable everywhere, so we may apply Rolle’s theorem on any interval. So, by Rolle’s theorem we know $f$ has at most two zeros (if there were three or more, say $x_1, x_2,$ and $x_3$ , then there must be distinct numbers $c_1$ and $c_2$ with $x_1 < c_1 < x_2 < c_2 < x_3$ such that $f'(c_1) = f'(c_2) = 0$ , but we know there is only one $c$ such that $f'(c) = 0$ ).
Furthermore, $f$ has at least two zeros since $f(-\pi) = \pi^2 + 1 > 0$ , $f(0) = -1 < 0$ , and $f(\pi) = \pi^2 + 1 > 0$ . Thus, by Bolzano’s theorem there are zeros between each of these points. We have that the number of zeros of $f$ is at most two and at least 2. Hence, the number of zeros must be exactly two $. \qquad \blacksquare$