The equation to be solve,
$$\tan 2x -\tan x=2$$
My Try :
$$\tan 2x=\frac{2\tan x}{1-\tan^2x}$$
$$\frac{2\tan x}{1-\tan^2x} -\tan x=2$$
$$\frac{2\tan x-\tan x(1-\tan ^2 x)}{1-\tan^2x} =2$$
$$2-2\tan^2x =2\tan x-\tan x+\tan^3x $$
$$\tan^3x+2\tan^2x+\tan x-2=0$$
$\tan x= t$
$$t^3+2t^2+t-2=0$$
now :?
The rational roots theorem gives $\pm(1,2)$ as candidate roots. None of these actually produce zeros. Your recourse is to either rely on the cubic formula or a numerical method, as qt commented.
If you want to explore a numerical solution by hand, you can hone in on the solution with a bit of analysis.
\begin{align} y&=t^3+2t^2+t-2 \\ \\ y'&=3t^2+4t+1 \\ &= (t+1)(3t+1) \end{align}
The function has critical points at $t=-1,-\frac{1}{3}$. We can determine that both $y(-1)<0$ and $y(-1/3)<0$. Given that this is a cubic polynomial, we conclude that $y<0$ (i.e. no zeros) for $t<0$.
Since $y(0)=-2$ and $y'>0$ for $t>0$, we know that there is a single zero for $t>0$. By inspection, it must lie within $(0,1)$. You could try Newton's method from here, if you want to churn out a numerical solution by hand.
Now you have to solve the polynomial. Unfortunately, there are no rational roots, so I suggest you to compute this
https://www.wolframalpha.com/input/?i=t%5E3%2B2t%5E2%2Bt-2%3D0
When you got the value of $t$, work again with tangent functions to find the values of $x$ that satisfy the first equation.
