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

Hint: In this question, we are given a trigonometric equation and we have been asked to solve it. Basically, we have been asked to find the value of $x$. Shift the constant term to the other side. Then, find the square root on both the sides.

Complete step-by-step answer:
We are given an equation and we have been asked to solve it. Let us see how it is to be done.
$ \Rightarrow {\tan ^2}x - 3 = 0$ …. (given)
Shifting the constant term to the other side,
$ \Rightarrow {\tan ^2}x = 3$
Now, we will square root both the sides.
$ \Rightarrow \sqrt {{{\tan }^2}x} = \sqrt 3 $
On simplifying, we get,
$ \Rightarrow \tan x = \pm \sqrt 3 $
Since $\tan x$ is both negative and positive in different quadrants, we will consider both the values.
But since there can be many values of $x$, how shall we find so many values?
In such a case, we will use the formula of general solution of trigonometric ratios.
We can write $\tan x = \pm \sqrt 3 $ as $\tan x = \tan \dfrac{\pi }{3}$.
The formula of general solution of $\tan x$ is –
If $\tan x = \tan \alpha $ then, $x = n\pi + \alpha $.
Using this formula in this question,
$ \Rightarrow x = n\pi + \dfrac{\pi }{3}$

Hence, $x = n\pi + \dfrac{\pi }{3}$

Note:
We could have also used the identity of ${a^2} - {b^2}$ to solve the given equation. Let us see how it is to be done.
We can also write ${\tan ^2}x - 3 = 0$ as ${\left( {\tan x} \right)^2} - {\left( {\sqrt 3 } \right)^2} = 0$.
Comparing with the identity, let $a = \tan x$ and $b = \sqrt 3 $.
The identity goes like this –
$ \Rightarrow {a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$
Hence, ${\left( {\tan x} \right)^2} - {\left( {\sqrt 3 } \right)^2} = \left( {\tan x + \sqrt 3 } \right)\left( {\tan x - \sqrt 3 } \right)$.
Putting each factor equal to $0$,$\tan x + \sqrt 3 = 0,\tan x - \sqrt 3 = 0$.
We will get, $\tan x = \sqrt 3 , - \sqrt 3 $
Now, we know that $\tan x$ has this value at numerous places. So, what value of x should be considered? To avoid this confusion, we use the general formula of x.
If $\tan x = \tan \alpha $ then, $x = n\pi + \alpha $.
Using this formula in this question,
$ \Rightarrow x = n\pi + \dfrac{\pi }{3}$
Hence, $x = n\pi + \dfrac{\pi }{3}$.