How do I solve a trigonometric equation?

Question

Thank you for looking in advance. It means alot :)

Here is my problem

Solve exactly for all values of $x\in[0,2\pi)$ such that $\tan^3x-3\tan x=0$.

I started by factoring out a $\tan x$. Then I set $\tan x = 0$ on one side, and on the other I ended up with $\tan x$ equals $\sqrt3$. I'm stuck after that... He provided the answer to the problem which is $x\in\{0,\pi,\pi/6,5\pi/6,7\pi/6, 11\pi/6\}$.

Answer 1

HINT:

Whenever you have $\tan^2x,\cos^2x,\sin^2x$ only in the equation, transform into $\cos2x$

$$\tan^2x=3\implies\cos2x=\frac{1-\tan^2x}{1+\tan^2x}=-\frac12=\cos\frac{2\pi}3$$

$$\implies 2x=2n\pi\pm\frac{2\pi}3\iff x=n\pi\pm\frac\pi3$$ where $n$ is any integer

Find suitable values of $n$ such that $x$ lies in $[0,2\pi)$

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Alternatively, $$\tan3x=\frac{3\tan x-\tan^3x}{1-3\tan^2x}$$

$$\implies\tan3x=0=\tan0\implies3x=m\pi$$ where $m$ is any intger

Find suitable values of $m$ such that $x$ lies in $[0,2\pi)$

Answer 2

Put $\tan x = t$.

Then $$\tan^3x-3\tan x=0 \iff t^2 - 3t = 0 \iff t(t^2 - 3) = 0$$ So for the first factor, find when $\tan x = 0$.

And for the second factor, we have a quadratic equation $t^2 - 3$ to solve, and once you find the two addition "zeroes", and then find when $\tan x = t$ for those solutions of $t$.

Alternatively, and more quickly, note that the second factor is a difference of squares: $$t^2 - 3 = 0 \iff t = \pm \sqrt 3$$