How can i resolve this trig equation?

Question

I don't know how to solve this kind of trig equation, could you help me?

$$3 \tan(\pi-x)+\tan^3(x)=0$$

Answer 1

tan(π - x) = - tan x

So equation becomes,

-3 tanx + $\tan^3x$ = 0

Taking tanx common,

tanx (-3 + $\tan^2x$) = 0

When tanx = 0

tanx = tan0 or tanx = tan nπ

x = 0 or nπ

When -3 + $\tan^2x$ = 0

$\tan^2x$ = 3

tanx = $\pm\sqrt3$

Case 1-

tanx = $\tan{\frac{π}{3}}$ or tanx = $\tan{\frac{π}{3} \pm πn}$

x = $\frac{π}{3}$ or $\frac{π}{3} \pm πn$

Case 2-

tanx = -$\tan{\frac{π}{3}}$

x = $\frac{2π}{3}$ and $\frac{5π}{3}$

Answer 2

You can recall that $\tan(\pi-x)=-\tan x$; so, if you set $y=\tan x$, the equation becomes $$ y^3-3y=0 $$ Can you solve it? Can you then solve $\tan x=y$ for the found values of $y$?