<s>[INST] $x_1$ and $x_2$ are the solutiton of $\frac{2\cdot \sin(x) \cdot \cos(2x)}{\cos (x) \cdot \sin (2x)} - 5 \tan(x) + 5 = 0$ $x_1$ and $x_2$ are the solutiton of $\frac{2\cdot \sin(x) \cdot \cos(2x)}{\cos (x) \cdot \sin (2x)} - 5 \tan(x) + 5 = 0$ 
then, $\tan(x_1 + x_2) = ....$
i can do it by doing it
$\dfrac{2\cdot \sin(x) \cdot (\cos^2x - \sin^2x)}{\cos (x) \cdot 2 \sin x \cos x} - 5 \tan(x) + 5 = 0$
leads to $(\sin x - \cos x)(\sin x + 6 \cos x) = 0$
but it's complicated, do you know the less complicated way to solve it?
 [/INST] Hint
Let $t=\tan(x)$ to make
$$2t \frac{1-t^2}{2t}-5t+5=0\implies t^2+5t-6=0$$
</s>