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Find the critical points of: $y = x + \cos(x)$

y' = 1 -\sin(x)

$\sin(x) = 1$ for stationary points.

$\therefore x = \frac{\pi}{2} + 2\pi k , k \in \mathbb{Z}$

I had trouble finding the value/s for $y$ since I wasn't sure of its period or how I would sub $x$'s period back into the original equation.

How do I find the values for $y$?

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    @ArturoMagidin: So, $y = x$ since the $cos(x)$ term is always =$0$at stationary points? I think that clears things up a lot, if I'm not missing anything else.2012-03-26

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You're essentially there: $y = x + \cos(x) = \frac\pi2 + 2\pi k + \cos(\frac\pi2 + 2\pi k) = \frac\pi2 + 2\pi k$. There are infinitely many $y$-values, one for each $k \in \mathbb{Z}$.

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$y(\frac{\pi}{2} + 2\pi k) = \frac{\pi}{2} + 2\pi k + \operatorname{cos}(\frac{\pi}{2} + 2\pi k) = \frac{\pi}{2} + 2\pi k $