Question

Find all the solutions of the given equations in the interval [0,2pi) tan^3x=tanx and cos 3x=-cos3x

Answer 1

$\tan^3x=\tan x$
$\tan^3x-\tan x=0$
$\tan x(\tan^2x-1)=0$
$\tan x(\tan x-1)(\tan x+1)=0$
$\begin{cases}\tan x=0\\\tan x-1=0\\\tan x+1=0\end{cases}$

$\tan x=\dfrac{\sin x}{\cos x}=0\implies \sin x=0\implies x=0,\pi$

$\tan x-1=0\iff\tan x=1\implies x=\dfrac\pi4,\dfrac{5\pi}4$

$\tan x+1=0\iff\tan x=-1\implies x=\dfrac{3\pi}4,\dfrac{7\pi}4$

- - -

$\cos3x=-\cos3x$
$2\cos3x=0$
$\cos3x=0\implies 3x=\dfrac\pi2,3x=\dfrac{3\pi}2\implies x=\dfrac\pi6,x=\dfrac\pi2$

This doesn't account for all the solutions, however; there are some values of $x$ that push $3x$ outside the interval $[0,2\pi)$, so let's take a few more:

$3x=\dfrac{5\pi}2\implies x=\dfrac{5\pi}6$
$3x=\dfrac{7\pi}2\implies x=\dfrac{7\pi}6$
$3x=\dfrac{9\pi}2\implies x=\dfrac{9\pi}6$
$3x=\dfrac{11\pi}2\implies x=\dfrac{11\pi}6$

We can stop there, since the next candidate gives

$3x=\dfrac{13\pi}2\implies x=\dfrac{13\pi}6>2\pi$