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

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?

• You want to figure out when $f'(x )> 0$. You have only figured out when it is equal to $0$. Set $\cos x > \frac{-1}2$ and find the desired range. Dec 2, 2017 at 23:41
• Please fix the title. Dec 2, 2017 at 23:42
• You have the correct inequality $2\cos x >-1$. But if you just sketch the cosine function, you will see because of periodicity, that there are many intervals where this is true. You got only one of the endpoints of the intervals, but with trigonometric functions you should in most cases expect infinite number of solutions. Dec 2, 2017 at 23:50
• Hence the $2pi*n$ but at $n=0$ the curve between $(-\frac{4\pi}{3}+2\pi n,\frac{4\pi}{3}+2\pi n)$ should always be increasing but I would like to see it. Dec 2, 2017 at 23:53
• wolframalpha.com/input/?i=x%2B2+sin(x) Dec 2, 2017 at 23:58 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 $$\bigcup_{k\in\mathbf Z}\Bigl(-\frac{2\pi}3+2k\pi,<\frac{2\pi}3+2k\pi\Bigr).$$