Alternative solutions for $\cos(x)=-\frac{1}{2}$ I have a function $f(x)=x+2\sin(x)$ and we are asked to find at which points the graph of $f$ has  Horizontal tangents. 
How we solve is to find where $f'=0$.  So $$f' = 1+ 2\cos x = 0 \implies \cos x = -\frac 12$$ and from this, to determine the solution set. 
I am familiar with unit circle and periodicity of $\sin,\cos$ functions so I found out that the solution set is $$x \in \left\{(2k+1)\pi \pm \frac{\pi}{3}:   k \in \mathbb{Z}\right\}$$
I know that is the usual way but is there any alternative solution -simpler if possible- to this problem? Especially for the people who does not the subject well?
 A: $f′(x)=0$ tells you where you can expect to find local maxima and minima (critical points), of the function, which occur when the tangent line to the curve is horizontal. 
You correctly found the derivative $f'(x) = 1 + 2\cos x$, and setting $f'=0$
you've solved for these $x$ quite well! Indeed, the solution set is given, as you found, by $$x \in \left\{(2k+1)\pi \pm \frac{\pi}{3}:   k \in \mathbb{Z}\right\}$$
So the horizontal tangent lines will be given by $$\left\{f(x): x \in \left\{(2k+1)\pi \pm \frac{\pi}{3}:   k \in \mathbb{Z}\right\}\right\}$$
Taking just one local maximum, when $x= \frac {2\pi}{3}$, we have the tangent line given by $$f\left(\frac {2\pi}3\right) = \frac{2\pi}{3} + \sqrt 3$$
We can compare the curve with one of its tangent lines:

A: One is 120. This can be found by noticing that $\cos(30)$ is $\frac{1}{2}$ and $\cos(x) = -\cos(x+90)$. Since $f(x) = \cos(x)$ has a period of 360, we can keep adding 360 to get solutions. Solutions are as follows: $\{\frac{2\pi}{3}, \frac{4\pi}{3}, \frac{8\pi}{3},\frac{10\pi}{3}, \cdots\}$
A: You can use the tan half angle substitution $t=\tan \frac{x}{2}$  to get
$$ \left. \cos x = -\frac{1}{2} \right\} \frac{1-t^2}{1+t^2} = \left. -\frac{1}{2} \right\} t = \pm \sqrt{3} $$
Then $$x = 2 \tan^{-1} t = \pm 2 \tan^{-1} \sqrt{3} = \pm 2 \frac{\pi}{3}$$
The periodicity means that you can add $k \pi$ angle to $x$ where $k=0,1\ldots \infty$
This is not a great application of the tan half angle method, but it is a different way to get the same result. 
Here are the rules of the Weierstrass substitution as it is called sometimes:

$$ \begin{align} t & =  \tan \frac{x}{2} \\ \sin x & = \frac{2 t}{1+t^2} \\ \cos x & = \frac{1-t^2}{1+t^2} \end{align} $$

