# Thread: Trigo-system

1. ## Trigo-system

Solve in :

2. Let $\tan x=a, \ \tan y=b$

Then, $\left\{\begin{array}{ll}a^2+b^2=6\\a^2+b^2=-6ab\end{array}\right.$

Now, let $s=a+b, \ p=ab$

$\Rightarrow\left\{\begin{array}{ll}s^2-2p=6\\s^2+4p=0\end{array}\right.\Rightarrow s=2, \ p=-1$ or $s=-2, \ p=-1$

If $s=2, \ p=-1$ then $a=1+\sqrt{2},b=1-\sqrt{2}$ or $a=1-\sqrt{2},b=1+\sqrt{2}$

If $s=-2,p=-1$ then $a=-1+\sqrt{2},b=-1-\sqrt{2}$ or $a=-1-\sqrt{2},b=-1+\sqrt{2}$

Now replace a and b and find x and y.

3. Hello, dhiab!

$\text{Solve in }\mathbb{R}^2\quad\begin{array}{cccc}\tan^2\!x + \tan^2\!y &=& 6 & {\color{blue}[1]} \\ \\[-4mm]\dfrac{\tan x}{\tan y} + \dfrac{\tan y}{\tan x} &=& \text{-}6 & {\color{blue}[2]} \end{array}$

From ${\color{blue}[2]}$ we have: . $\frac{\tan^2\!x + \tan^2\!y}{\tan x\tan y} \:=\:-6$

Substitute ${\color{blue}[1]}$: . $\frac{6}{\tan x\tan y} \:=\:-6 \quad\Rightarrow\quad \tan x\tan y \:=\:-1 \quad\Rightarrow\quad \tan y \:=\:\frac{-1}{\tan x}\;\;{\color{blue}[3]}$

Substitute into ${\color{blue}[1]}:\quad \tan^2\!x + \left(\frac{-1}{\tan x}\right)^2 \:=\:6 \quad\Rightarrow\quad \tan^2\!x + \frac{1}{\tan^2\!x} \:=\:6$

. . $\tan^4\!x + 1 \:=\:6\tan^2\!x \quad\Rightarrow\quad \tan^4\!x - 6\tan^2\!x + 1 \:=\:0$ . . . a quadratic in $\tan^2\!x$

Quadratic Formula: . $\tan^2\!x \;=\;\frac{6 \pm\sqrt{32}}{2} \:=\:3 \pm 2\sqrt{2} \:=\:(1 \pm\sqrt{2})^2$

. . Hence: . $\tan x \:=\:\pm(1 \pm\sqrt{2})$

Substitute into ${\color{blue}[3]}\!:\quad \tan y \:=\:\frac{-1}{\pm(1\pm\sqrt{2})}$

. . Hence: . $\tan y \;=\;\pm(1\mp\sqrt{2})$

Therefore: . $\begin{array}{ccc}x &=& \arctan\left[\pm(1 \pm\sqrt{2})\right] \\ \\[-3mm]
y &=& \arctan\left[\pm(1\mp\sqrt{2})\right]\end{array}$

I'll let you sort out the answers and determine if any are extraneous.

4. Hell,: red dog and Soroban ;Thanks for solution.
I'have solution when can calculate x and y.
I'have this identity :

For : :

I'have this equation :

The solution is :

coclusion I :
same solution for
For :
I'have this equation :
The solution is :
Conclusion the solution of system is :