Phương trình đại số có nghiệm lượng giác //Algebraic Equation with Trigonometric Solution

 We are showing here what we have left from a post from yesterday. I was helped for this in the AOPS Forum.

               The roots of the equation

$$t^3-3\sqrt{3}\cdot t^2-3\cdot t+\sqrt{3}=0 \tag{E}  $$

             are  $\tan20^{\circ},\;\;\tan80^{\circ},\;\;\tan140^{\circ}$

Adapted Solution by CiP

$(E)\Leftrightarrow \sqrt{3}\cdot (1-3t^2)=3t-t^3 \Leftrightarrow \sqrt{3}=\frac{3t-t^3}{1-3t^2}\;\underset{t=\tan x}{=}\frac{3\cdot \tan x-\tan^3x}{1-3\cdot \tan^2x}=\tan (3x)$

so, taking the equality of the extreme terms,  $3\cdot x=60^{\circ}+k\cdot 180^{\circ},\;\;k\in \mathbb{Z}$, hence  $x_k=20^{\circ}+k\cdot 60^{\circ}$. Returning to the variable  $t$,  we have the roots

$t_1=\tan( 20^{\circ}+3k\cdot 60^{\circ})=\tan (20^{\circ}+k\cdot 180^{\circ})=\tan20^{\circ},$

$t_2=\tan(20^{\circ}+(3k+1)\cdot 60^{\circ})=\tan(80^{\circ}+k\cdot 180^{\circ})=\tan80^{\circ}$

$t_3=\tan(20^{\circ}+(3k+2)\cdot 60^{\circ})=\tan(140^{\circ}+k\cdot 180^{\circ})=\tan140^{\circ}$.

$\blacksquare$

          Remark Cip  Vieta's formulas for the roots of the equation (E) lead to the equalities

$\tan20^{\circ}-\tan40^{\circ}+\tan80^{\circ}=3\sqrt{3}$

$\tan20^{\circ}\cdot \tan 40^{\circ}+\tan 40^{\circ}\cdot \tan 80^{\circ}-\tan 20^{\circ}\cdot \tan 80^{\circ}=3$

$\tan20^{\circ}\cdot \tan 40^{\circ}\cdot \tan80^{\circ}=\sqrt{3}$

<end Rem>