How to find all solutions of $\tan(x) = 2 + \tan(3x)$ without a calculator?

Question

Find all solutions of the equation $\tan(x) = 2 + \tan(3x)$ where $0<x<2\pi$.

By replacing $\tan(3x)$ with $\dfrac{\tan(2x) + \tan(x)}{1-\tan(2x)\tan(x)}$ I've gotten to $\tan^3 (x) - 3 \tan^2 (x) + \tan(x) + 1 =0$.

I am not sure how to proceed from there without the use of a calculator.

Answer 1

Put $t=\tan(x)$, so that your equation become: $t^3-3t^2+t+1=0$. Since $t^3-3t^2+t+1=(t-1)(t^2-2t-1)$, it follows that the solutions are $t=1$, $t=1\pm\sqrt 2$, from which

$x\equiv\arctan(1)\equiv\frac\pi 4\pmod \pi$,

$x\equiv\arctan(1+\sqrt 2)\equiv\frac{3\pi}8\pmod\pi$,

$x\equiv\arctan(1-\sqrt 2)\equiv-\frac\pi8\pmod\pi$.

Answer 2

Let $\tan x=t$. Then, we have $$t=2+\frac{3t-t^3}{1-3t^2}\Rightarrow (t-2)(1-3t^2)=3t-t^3$$$$\Rightarrow t^3-3t^2+t+1\Rightarrow (t-1)(t^2-2t-1)=0.$$ Hence, we have $$\tan x=t=1,1\pm\sqrt 2\Rightarrow x=\frac{\pi}{4}+n\pi,\frac{3}{8}\pi+n\pi,-\frac{\pi}{8}+n\pi$$ where $n\in\mathbb Z$.

Answer 3

$t=\tan x=1$ is easy to deal with

$$t^2-2t-1=0\implies -2t=1-t^2\implies\frac{2t}{1-t^2}=-1$$

$$\tan2x=-1=-\tan\frac\pi4=\tan\left(-\frac\pi4\right)$$

$$2x=n\pi-\frac\pi4$$ where $n$ is any integer