Question

How we do this one there more than one answer

Answer 1

Given;

`\tan^2x-(1)/(3)=0,[0,2\pi)`

Thus;

`\begin{gathered} \tan^2x=(1)/(3) \\ \\ \tan x={\pm}\sqrt{(1)/(3)} \end{gathered}`

Thus;

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

So, the solution is;

`x=(\pi)/(6),(7\pi)/(6),(5\pi)/(6),(11\pi)/(6)`