$x_1$ 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?

Convert all to $\tan$:
$$\frac{2 \sin(x) \cos(2x)}{\cos(x) \sin(2x)} - 5 \tan(x) +5 = \frac{2 \tan(x)}{\tan(2x)} - 5 \tan(x) +5 = (1 - \tan^2(x)) - 5 \tan(x) + 5$$
$$ \Longrightarrow \tan^2(x) + 5 \tan(x) - 6 = 0$$
Now we have:
$$\tan(x_1 + x_2) = \frac{\tan(x_1) + \tan(x_2)}{1 - \tan(x_1)\tan(x_2)} = \frac{-b/a}{1-c/a} = \frac{-5/1}{1-(-6/1)} = \frac{-5}{1+6} = -\frac{5}{7}$$
Without need to compute the roots!
