5
$\begingroup$

$$\tan x-\tan(2x)=2\sqrt{3}$$


TRY #1

$$\begin{align*} \tan x-\tan(2x)=2\sqrt{3}&\implies\tan x=2\sqrt{3}+\tan{2x}\\ &\implies \tan^2x=\tan^2(2 x)+4 \sqrt{3} \tan(2 x)+12\\ &\implies\tan^2x=(\frac{2\tan x}{1-\tan^2 x})^2+4\sqrt{3}\frac{2\tan x}{1-\tan^2x}+12 \end{align*}$$

but this will give me an equation with $\tan^4$ which needs quartic formula, too difficult!!


TRY #2

$$\begin{align*} \tan x-\tan(2x)=2\sqrt{3} &\implies \frac{\sin x}{\cos x}-\frac{\sin 2x}{\cos 2x}=2\sqrt3 \\ &\implies\frac{\sin x\cos 2x-\sin 2x\cos x}{\cos x\cos 2x}=2\sqrt{3}\\ &\implies\frac{-\sin x}{\cos x\cos 2x}=2\sqrt{3}\\ &\implies\frac{-\sin x-2\sqrt{3}\cos x\cos 2x}{1}=0 \end{align*} $$ then i can't!!

can anyone help me?

$\endgroup$
  • 2
    $\begingroup$ Knowing that $\tan(\pi/3) = \sqrt{3}$ and $\tan(2\pi/3) = -\sqrt{3}$ might help. $\endgroup$ – Winther Jan 6 '15 at 20:54
  • 2
    $\begingroup$ Don't use $\therefore$, especially not three of them. It's unreadable. I fixed it for you. $\endgroup$ – Emily Jan 6 '15 at 20:56
  • $\begingroup$ @Winther i know that, but still couldn't! $\endgroup$ – user205150 Jan 6 '15 at 21:00
  • $\begingroup$ @Winther also can we solve it using weirstrauss substitution $t=\tan x/2$ ? $\endgroup$ – user205150 Jan 6 '15 at 21:04
3
$\begingroup$

How about

$$\tan x-\frac{2\tan x}{1-\tan^2 x}=2\sqrt 3$$

Removing the fraction will give you a cubic equation. At least that is easier than the quartic!

$\endgroup$
  • $\begingroup$ i don't know how to solve cubics, only quadratics and simpler! $\endgroup$ – user205150 Jan 6 '15 at 21:01
  • 3
    $\begingroup$ @user205150 Hint: You know one solution of the qubic! $\tan(x) = \sqrt{3}$. Now you can write the cubic as $(T-\sqrt{3})(T^2+aT+b) = 0$. Solve for $a,b$ and then you only need to solve a second order equation. $\endgroup$ – Winther Jan 6 '15 at 21:02
3
$\begingroup$

let $u = \tan x, \tan 2x = \dfrac{2u}{1-u^2}$ your equation becomes $$u - \dfrac{2u}{1-u^2} = 2\sqrt 3 $$ which can be simplified $$f(u) = u^3 - 2\sqrt 3 u^2 + u + 2\sqrt 3 = 0 $$ i don't see any simple roots for this. we know that $f(0) = 2\sqrt 3$ and $f(-1) = -2$ so there is at least one $-1 < u < 0, f(u) = 0$

edit: thanks to user winther, we can factor it.

$$u^3 - 2\sqrt 3 u^2 + u + 2\sqrt 3 = (u-\sqrt 3)(u^2 -\sqrt 3 u -2) $$

$\endgroup$
  • 2
    $\begingroup$ Try $u=\sqrt{3}$. $\endgroup$ – Winther Jan 6 '15 at 21:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.