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.

• Why did your attempts fail? What you tried is what I'd try, that is $\sin (x)=0\lor \sin (x)\neq 0$. – Git Gud Aug 18 '13 at 18:34
• I reached the fact that $x \in [-\pi,-2\pi/3]\cup[-\pi/3,\pi/3]\cup[2\pi/3,\pi]$ which isn't the solution set according to wolfram|alpha. – user7610 Aug 18 '13 at 18:38
• You should seriously consider multiplying by a certain trigonometric function, and dividing by another one, where permitted and being careful about signs. – dfeuer Aug 18 '13 at 18:40

$|\tan(x)|\le2\sin(x)$

$|\frac{\sin(x)}{\cos(x)}|\le2\sin(x)$

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 $x\in[0,\frac{\pi}{3}]\cup[\frac{2\pi}{3},\pi]\cup\{-\pi\}$

• This looks promising. But to elaborate on my second question, do I also need to demonstrate that $x\in\{-\pi\}\cup[0,\frac{\pi}{3}]\cup[\frac{2\pi}{3},\pi]$ implies $|\tan(x)|\le2\sin(x)$? or is each step here an equivalence? – user7610 Aug 18 '13 at 19:32
• I've started by assuming the inequality and then found a simpler expression where we can find the regions where it is true. So it is demonstrated. Each step is an equivalence. – user71352 Aug 18 '13 at 19:35
• I don't understand why there is $\cup \{\pi\}$... isn't the case when $x = -\pi$ the same case of $x = \pi$? – user67133 Aug 18 '13 at 19:50
• It does represent the same case as $x=\pi$. I put it there because I don't recall OP saying we mod out by $2\pi$. If we did do that then we would associate them. – user71352 Aug 18 '13 at 19:58

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.