How do you solve \[3{\tan ^2}x - 1 = 0\]?

Hint: In this question we have to solve the equation for $ x $ , first take all constant terms to one side, take out the square root, then by using trigonometric ratios we will get the equation in the form of $ \tan x = \tan \theta $ , we know that the general solution of $ \tan x = \tan \theta $ is $ x = n\pi + \theta $ , where $ n \in Z $ $ $ (i.e., $ n = 0 $ , $ \pm 1 $ , $ \pm 2 $ , $ \pm 3 $ ,…….) and $ \theta \in \left( {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right) $ now by substituting the values we will get the required result.

Complete step by step solution:
Given equation is \[3{\tan ^2}x - 1 = 0\],
First add 1 to both sides of the equation we get,
\[ \Rightarrow 3{\tan ^2}x - 1 + 1 = 0 + 1\],
Now simplifying we get,
\[ \Rightarrow 3{\tan ^2}x = 1\],
Now first divide both sides with 3 we get,
$ \Rightarrow \dfrac{{3{{\tan }^2}x}}{3} = \dfrac{1}{3} $ ,
Now simplifying we get,
$ \Rightarrow {\tan ^2}x = \dfrac{1}{3} $ ,
Now taking out the square root we get,
$ \Rightarrow \tan x = \pm \sqrt {\dfrac{1}{3}} $ ,
Again simplifying we get,
$ \Rightarrow \tan x = \pm \dfrac{1}{{\sqrt 3 }} $ ,
Now the values for $ \tan x $ are $ \dfrac{1}{{\sqrt 3 }} $ and $ - \dfrac{1}{{\sqrt 3 }} $ ,
Take first value i.e..,
$ \Rightarrow \tan x = \dfrac{1}{{\sqrt 3 }} $ ,
Now using trigonometric ratio table we get, $ \tan \dfrac{\pi }{6} = \dfrac{1}{{\sqrt 3 }} $ , so write $ \dfrac{1}{{\sqrt 3 }} $ as $ \tan \dfrac{\pi }{6} $ , we get,
$ \Rightarrow \tan x = \tan \dfrac{\pi }{6} $ ,
So, now using the fact that the general solution of $ \tan x = \tan \theta $ is $ x = n\pi + \theta $ , where $ n \in Z $ $ $ (i.e., $ n = 0 $ , $ \pm 1 $ , $ \pm 2 $ , $ \pm 3 $ ,…….) and $ \theta \in \left( {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right) $ ,
Now, by substituting the values, here $ \theta = \dfrac{\pi }{6} $ , in the general solution we get,
$ \Rightarrow x = n\pi + \dfrac{\pi }{6} $ ,
Take second value i.e..,
$ \Rightarrow \tan x = - \dfrac{1}{{\sqrt 3 }} $ ,
Now using trigonometric ratio table we get, \[\tan \left( { - \dfrac{\pi }{6}} \right) = - \dfrac{1}{{\sqrt 3 }}\], so write $ - \dfrac{1}{{\sqrt 3 }} $ as $ \tan \left( { - \dfrac{\pi }{6}} \right) $ , we get,
$ \Rightarrow \tan x = \tan \left( { - \dfrac{\pi }{6}} \right) $ ,
Since tan will have negative value in second and fourth quadrant, we can write the angle as,
$ \Rightarrow \tan x = \tan \left( {\pi - \dfrac{\pi }{6}} \right) $ ,
Now simplifying we get,
$ \Rightarrow \tan x = \tan \left( {\dfrac{{5\pi }}{6}} \right) $ ,
So, now using the fact that the general solution of $ \tan x = \tan \theta $ is $ x = n\pi + \theta $ , where $ n \in Z $ $ $ (i.e., $ n = 0 $ , $ \pm 1 $ , $ \pm 2 $ , $ \pm 3 $ ,…….) and $ \theta \in \left( {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right) $ ,
Now, by substituting the values, here $ \theta = \dfrac{{5\pi }}{6} $ , in the general solution we get,
$ \Rightarrow x = n\pi + \dfrac{{5\pi }}{6} $ ,
So, the general solutions are $ x = n\pi + \dfrac{\pi }{6} $ and $ x = n\pi + \dfrac{{5\pi }}{6} $ .

$ \therefore $ The general solution for the given function i.e., \[3{\tan ^2}x - 1 = 0\]will be equal to $ x = n\pi + \dfrac{\pi }{6} $ and $ x = n\pi + \dfrac{{5\pi }}{6} $ .

Note: Trigonometric equations are those equations that involve the trigonometric functions as a variable. Principal solutions are those solutions that lie in the interval $ \left[ {0,2\pi } \right] $ of such trigonometric equations, and trigonometric equation will also have general solution expressing all the values which would satisfy the given equation and it is expressed in a generalised form in terms of ‘n’.