Solving for the principal value of a trigonometric equation

Question

The question is as follows:

$$3{\tan}^2(2x)-1=0.$$ Solve for $x$.

What steps should I take to solve? The squaring is really throwing me off.

Answer 1

Well, $$3 \tan^2(2x) - 1 = 0 \\ \tan^2(2x) = \frac{1}{3} \\ \tan(2x) = \pm \frac{1}{\sqrt{3}}.$$ Can you take it from there?

Answer 2

$$\frac{\tan^22x}1=\frac13$$

Applying by Componendo and dividendo,

$$\frac{1-\tan^22x}{1+\tan^22x}=\frac{3-1}{3+1}$$

$$\implies \cos4x=\frac12=\cos\frac\pi3\implies4x=2n\pi\pm\frac\pi3$$ where $n$ is any integer