[tex]\tan^3x=\tan x[/tex]
[tex]\tan^3x-\tan x=0[/tex]
[tex]\tan x(\tan^2x-1)=0[/tex]
[tex]\tan x(\tan x-1)(\tan x+1)=0[/tex]
[tex]\begin{cases}\tan x=0\\\tan x-1=0\\\tan x+1=0\end{cases}[/tex]
[tex]\tan x=\dfrac{\sin x}{\cos x}=0\implies \sin x=0\implies x=0,\pi[/tex]
[tex]\tan x-1=0\iff\tan x=1\implies x=\dfrac\pi4,\dfrac{5\pi}4[/tex]
[tex]\tan x+1=0\iff\tan x=-1\implies x=\dfrac{3\pi}4,\dfrac{7\pi}4[/tex]
- - -
[tex]\cos3x=-\cos3x[/tex]
[tex]2\cos3x=0[/tex]
[tex]\cos3x=0\implies 3x=\dfrac\pi2,3x=\dfrac{3\pi}2\implies x=\dfrac\pi6,x=\dfrac\pi2[/tex]
This doesn't account for all the solutions, however; there are some values of [tex]x[/tex] that push [tex]3x[/tex] outside the interval [tex][0,2\pi)[/tex], so let's take a few more:
[tex]3x=\dfrac{5\pi}2\implies x=\dfrac{5\pi}6<2\pi[/tex]
[tex]3x=\dfrac{7\pi}2\implies x=\dfrac{7\pi}6<2\pi[/tex]
[tex]3x=\dfrac{9\pi}2\implies x=\dfrac{9\pi}6<2\pi[/tex]
[tex]3x=\dfrac{11\pi}2\implies x=\dfrac{11\pi}6<2\pi[/tex]
We can stop there, since the next candidate gives
[tex]3x=\dfrac{13\pi}2\implies x=\dfrac{13\pi}6>2\pi[/tex]
