Finding all $x$ such that $|\tan x | \leq 2\sin x$ I need to find all real numbers $x$ that satisfy: $$|\tan x | \leq 2\sin x \text{ and } x \in [ -\pi, \pi]$$ in terms of unions of intervals.
I know it's equivalent to: $-2\sin x \leq \tan x \leq 2 \sin x $
I tried dividing into cases where $\sin x = 0 $ or $\sin x \neq0$   .
and also, $\cos x \gt 0 $ or $\cos x \lt 0 $. But alas, my attempts to solve this failed. 
Perhaps I'm missing something?
In addition, I'm struggling with these types of questions (finding solution sets) as it's hard for me to see whether the attempted solution shows a double inclusion or a single one. In other words whether each step of inference is an equivalence or just an implication.
Thanks.
 A: When $x=0$, then $\tan x = 2\sin x$.  As soon as $x$ is a little bit bigger than $0$, then $2\sin x$ is bigger than $\tan x$, since $2\sin x$ is growing faster than is $\tan x$ at the point where $x=0$.  But as $x\uparrow\pi/2$, $\tan x$ goes up to $\infty$ and $2\sin x$ does not, so $\tan x$ has to overtake $2\sin x$.  So the question is, at what point does $\tan x$ overtake $2\sin x$, and that has to happen when they're both in the same place.  In other words, we need to solve
$$
\tan x = 2\sin x.
$$
Dividing both sides by $\sin x$ (so this applies only at points where $\sin x\ne0$), we get
$$
\frac{1}{\cos x} = 2,
$$
so $\cos x=1/2$.  And as you learned in trigonometry, that happens when $x=\pi/3$.
Thus the proposed inequality holds for nonnegative values of $x$ satisfying $0\le x\le \pi/3$.  For negative values of $x$, just recall that sine and tangent are odd functions.
A: $|\tan(x)|\le2\sin(x)$
$|\frac{\sin(x)}{\cos(x)}|\le2\sin(x)$
It only makes sense to consider $x\in[0,\pi]\cup\{-\pi\}$ since $0\le|\tan(x)|$ and $\sin(x)\ge0$ for these values. For $x\in(0,\pi)$ this says
$\frac{\sin(x)}{|\cos(x)|}\le2\sin(x)$
$\frac{1}{2}\le|\cos(x)|$ since $\sin(x)>0$ here.
The above gives $x\in[0,\frac{\pi}{3}]\cup[\frac{2\pi}{3},\pi]\cup\{-\pi\}$
