How to prove that $x^2+10\cos x=0$ has exactly 2 positive solutions? I located the two roots in $[1,\pi]$ and $[\pi,4]$ but i cannot prove that there is not a third one. I tried using Rolle's theorem. Any ideas?
 A: The derivative is $2x-10\sin(x)$, from which we learn that the original function looks roughly like $x^2$, in the sense that it has a minimum and is monotonic (on $[0,\infty)$) on either side of the minimum. This is because the derivative has one zero in the positive numbers^. Thus it can only cross the $x-axis at most twice.
^To prove this claim, let $g(x)=2x-10\sin(x)$. We see that $2x-10\leq g(x)\leq 2x+10$, so any zeros must occur before $x=5$. We observe that there is one zero at approximately $2.5$, so we want to know where the others are relative to that point. Using our favorite approximation technique, we can find out that this zero is greater than $2.5$. Since $2.5+\pi>5$ we see that at the next local minimum the function is greater than $0$. It follows that the function is positive between the zero near $2.5$ and $5$. Likewise, we reach $x=0$ to the left before completing a full period. This is before the next local max to the left, so the function was monotonic decreasing between $0$ and the local minimum, meaning it can only cross the $x$ axis once. But it crosses the $x$ axis at $0$, so there are no other zeros, since we've already observed the zero in the stretch where the function is increasing.
