<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] The equation is $ \frac{2\tan x}{\tan 2x} - 5 \tan x + 5 = 0$, as one sees from rearranging the first term. The double angle formula for $\tan$ gives $\frac{2 \tan x }{\tan 2x} = 1 - \tan^2 x$ and therefore $(1 - \tan^2 x) - 5 \tan x + 5 = 0$, which simplifies to $\tan^2 x + 5 \tan x - 6 = 0$. This finally gives $\tan x = -6$ or $\tan x = +1$.
Finally, using the addition formula, we get $\tan(x_1 + x_2) = \frac{-6 + 1}{1 - (-6)\cdot(+1)} = \frac{-5}{7}$.
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