Given that there is at least one $x$ such that $\cos x=x$ for $x\in (0,\pi/2)$, how do I show there are at most finitely many such $x\in (0,\pi/2)$? I want to use properties of compactness (maybe a subsequence converging?) but I'm not sure how to go forward.
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You will need to use some property specific to $\cos x$, since a general continuous function may have infinitely many fixed points in an interval. Ex: $g(x) = (x-\pi/4)\cos(1/(x-\pi/4)) + x$ is continuous on $(0,\pi/2)$ and has infinitely many fixed points there. For your problem we can use the fact that $\cos x$ is complex analytic on $(0,\pi/2)$ and not identically equal to $x$ there, so the equation $\cos x = x$ must only have finitely many solutions (or else it violates the identity theorem). You've tagged this [tag:real-analysis] though, so this might not be the kind of argument you seek. – Antonio Vargas Oct 27 '15 at 20:45
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Yes, another approach would be helpful. – Moz Oct 27 '15 at 21:37
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On reflection we can just use the fact that $\cos x$ is real analytic, and use the real analytic version of the identity theorem. – Antonio Vargas Oct 28 '15 at 02:36