Solve $2\tan{2x}\leq3\tan{x}.$

Question

Problem:

$$\text{Solve} \quad 2\tan{2x}\leq3\tan{x}.$$

A problem of this character will yield 5 points on an exam. However, having the correct answer does not suffice to get all the 5 points. Full stringency and mathematical accuracy, on top of a correct final answer, warrants full house. I've decided to present a (partial) solution here and I want you to help me really to comb through it and search for possible logical loopholes.

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Solution attempt:

Application of the double angle formula for $\tan{x}$ on LHS yields:

$$\text{LHS}=2\cdot\frac{\sin{2x}}{\cos{2x}}=2\cdot\frac{2\sin{x}\cos{x}}{\cos^2{x}-\sin^2{x}}=[\text{Divide by} \ \cos^2{x}\neq0]=\frac{4\tan{x}}{1-\tan^2{x}}.$$

Setting $t=\tan{x}$ and moving the RHS over and subtracting gives the equivalent inequality:

$$\frac{4t}{1-t^2}-3t=\frac{4t-3t(1-t^2)}{1-t^2}=\frac{t(1+3t^2)}{(1+t)(1-t)}=p(t)\leq0.$$

Since the factor $(1+3t^2)>0, \ \forall x\in \mathbb{R}, $ it suffices to examine the signs of the factors $t, \ (1+t), \ (1-t)$ and the entire expression that I denoted $p(t).$ The following table emerges:

\begin{array} {|l|cr} t= & -\infty & -1 & \ & 0 & \ & 1 & +\infty\\ \hline 1+t & - & 0 & + & & + & & +\\ \hline t & - & & - & 0 & + & & +\\ \hline 1-t & + & & + & & + & 0 & -\\ \hline p(t)& + & \varnothing & - & 0 & + & \varnothing & -\\ \end{array}

This indicates that the solutionset of $p(t)\leq0$ is given by $t\in(-1,0]\cup(1,\infty).$ Hereafter I'm stuck, I don't know how to revert to $x$. How do I do this in an effective way?

Answer 1

The goal is to find conditions on $x$ such that $\tan(x)$ will be in the indicated intervals. Recall that if $x\in \left[ -\frac{\pi}{2},\frac{\pi}{2}\right] =: \mathcal{D}$, then $$\tan(x) = y \iff \arctan(y) = x.$$ It isn't too difficult to see that $$ \arctan(-1) = -\frac{\pi}{4}, \qquad\text{and}\qquad \arctan(0) = 0.$$ Since $\arctan$ is increasing and continuous on its domain, it follows that $\tan(x) \in (-1,0]$ and $x\in\mathcal{D}$ if and only if $$ x \in (\arctan(-1),\arctan(0)] = \left( -\frac{\pi}{4}, 0\right].$$ By a similar argument, we conclude that $\tan(x) \in (1,\infty)$ and $x\in\mathcal{D}$ if and only if $$ x \in \left(\arctan(1),\lim_{y\to\infty} \arctan(y)\right) = \left( \frac{\pi}{4}, \frac{\pi}{2} \right).$$ As the tangent function is periodic with fundamental period equal to $\pi$, it follows that if $x$ satisfies the given inequality, then so to will $x + k\pi$ for any $k \in\mathbb{Z}$. Therefore the complete set of solutions is given by $$ \bigcup_{k\in\mathbb{Z}} \left[ \left( -\frac{\pi}{4} + k\pi, k\pi\right] \cup \left( \frac{\pi}{4} + k\pi, \frac{\pi}{2}+k\pi \right) \right].$$

Answer 2

Duplication formula for tangent

$\dfrac{4 \tan x}{1-\tan ^2 x}\leq 3 \tan x$

Bring all in the LHS

$\dfrac{4 \tan x}{1-\tan ^2 x}-3 \tan x\leq 0$

add together

$-\dfrac{\tan x \left(3 \tan ^2 x+1\right)}{\tan ^2 x-1}\leq 0$

which is better as

$\dfrac{\tan x \left(3 \tan ^2 x+1\right)}{\tan ^2 x-1}\geq 0$

The parenthesis $\left(3 \tan ^2 x+1\right)$ is positive for any $x$ because is the sum of two squares so we need to see where

$\tan x \geq 0$ verified in $0\leq x <\dfrac{\pi}{2}\lor \pi\leq x <\dfrac{3\pi}{2}$

and $\tan ^2 x-1>0$ verified when $\tan x<-1\lor \tan x>1$

which is $\dfrac{\pi }{4}<x<\dfrac{\pi }{2}\lor \dfrac{\pi }{2}<x<\dfrac{3 \pi }{4}\lor \dfrac{5 \pi }{4}<x<\dfrac{3 \pi }{2}\lor \dfrac{3 \pi }{2}<x<\dfrac{7 \pi }{4}$

The solution is then

$\dfrac{\pi }{4}<x<\dfrac{\pi }{2}\lor \dfrac{3 \pi }{4}<x\leq\pi \lor \dfrac{5 \pi }{4}<x<\dfrac{3 \pi }{2}\lor \dfrac{7 \pi }{4}<x\leq2 \pi$

Answer 3

We need to solve $$\frac{4\tan{x}}{1-\tan^2x}\leq3\tan{x}$$ or $$\tan{x}\left(\frac{4}{1-\tan^2x}-3\right)\leq0$$ or $$\frac{\tan{x}(3\tan^2x+1)}{(1-\tan{x})(1+\tan{x})}\leq0,$$ which by the intervals method gives $$\tan{x}<-1$$ or $$0\leq\tan{x}<1$$ and we got the answer: $$\left[\pi k\leq x<\frac{\pi}{4}+\pi k\right)\cup\left(-\frac{\pi}{2}+\pi k,-\frac{\pi}{4}+\pi k\right),$$ where $k\in\mathbb Z$.

For example, the inequality $\tan{x}<-1$ we can solve by the following way.

Draw the trigonometric circle (the circle $x^2+y^2=1$) and the tangent to the circle in the point $B(1,0)$.

This line names a tangents-axis.

Let $A(1,a)$ placed in the tangents-axis.

Thus, easy to see that $\tan\measuredangle AOB=a$ and since we need to solve $\tan{x}<-1$,

we got $-\frac{\pi}{2}<x<-\frac{\pi}{4}$ (see on the tangents-axis).

Since the period of $\tan$ is equal to $\pi$, finally we obtain: $$-\frac{\pi}{2}+\pi k<x<-\frac{\pi}{4}+\pi k,$$ where $k\in\mathbb Z$.

By the similar way we can solve $0\leq\tan{x}<1$.