5

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

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?

Emily
  • 35,688
  • 6
  • 93
  • 141
user205150
  • 51

2 Answers2

3

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!

Rory Daulton
  • 32,288
  • i don't know how to solve cubics, only quadratics and simpler! – user205150 Jan 06 '15 at 21:01
  • 3
    @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. – Winther Jan 06 '15 at 21:02
3

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.

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

abel
  • 29,170