Question

how we do this one there more than one answer

Answer 1

Solution:

Given;

$\tan^2x-\frac{1}{3}=0,[0,2\pi)$

Thus;

$\begin{gathered} \tan^2x=\frac{1}{3} \\ \\ \tan x={\pm}\sqrt{\frac{1}{3}} \end{gathered}$

Thus;

$\begin{gathered} x=\tan^{-1}(\frac{1}{3}) \\ \\ x=\frac{\pi}{6},\frac{7\pi}{6} \\ \\ x=\tan^{-1}(-\frac{1}{3}) \\ \\ x=\frac{5\pi}{6},\frac{11\pi}{6} \end{gathered}$

So, the solution is;

$x=\frac{\pi}{6},\frac{7\pi}{6},\frac{5\pi}{6},\frac{11\pi}{6}$