Finding the interval for which $f(x) =x+2\sin(x)$ is increasing

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I have a the function $f(x)=x+2\sin(x)$ and I want to find the increasing interval.

So I find the derivative when it's larger than 0.

Hence $f'(x)>0$ when $2\cos(x)>-1$.

So by figuring when $f'(x) = 0$ and got it to

$\cos(x)=-\frac{1}{2}$ so $x=\frac{4\pi}{3}$

according to the formula the increasing interval is between $(-\frac{4\pi}{3}+2\pi n,\frac{4\pi}{3}+2\pi n)$

I don't really understand how that's possible. Shouldn't it during some instance decrease within the interval? Is there some program where I could visualise the increase between these points?

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2 Answers

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Perhaps this diagram will help:

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You have to solve $\;1+2\cos x>0\iff\cos x>_\frac12$.

Solve it first on $[-\pi,\pi]$. Making a sketch on the unit circle leads to $$-\frac{2\pi}3<x<\frac{2\pi}3,$$ whence the general solution $$\bigcup_{k\in\mathbf Z}\Bigl(-\frac{2\pi}3+2k\pi,<\frac{2\pi}3+2k\pi\Bigr).$$

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