How do you solve $3{{\cot }^{2}}x-1=0$ between the interval $0\le x\le 2\pi $?

Hint: We first have to factorise the trigonometric function on the left hand side. For this we need to use the algebraic identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$. After factoring, we will get two equations. According to the interval given in the question, $0\le x\le 2\pi $, the solution will lie in all of the four quadrants. Using the principle value of the solution, we can arrange the value according to all of the four quadrants. Thus, we will obtain four solutions to the given equation.

Complete step by step answer:
The trigonometric equation given in the question is
$3{{\cot }^{2}}x-1=0$
And the interval given to us is $0\le x\le 2\pi $.
This means that the solution of the above equation will lie in all of the four quadrants, and therefore there will be four solutions.
Dividing both the sides of the above equation by $3$ we get
$\Rightarrow {{\cot }^{2}}x-\dfrac{1}{3}=0$
Now, writing $\dfrac{1}{3}={{\left( \sqrt{\dfrac{1}{3}} \right)}^{2}}$, we get
\[\begin{align}
  & \Rightarrow {{\cot }^{2}}x-{{\left( \sqrt{\dfrac{1}{3}} \right)}^{2}}=0 \\
 & \Rightarrow {{\left( \cot x \right)}^{2}}-{{\left( \sqrt{\dfrac{1}{3}} \right)}^{2}}=0.......(i) \\
\end{align}\]
Now, we know the algebraic identity
${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$
Comparing this identity with the equation (i), we have $a=\cot x$ and $b=\sqrt{\dfrac{1}{3}}$. So the equation (i) can be written as
\[\Rightarrow \left( \cot x+\sqrt{\dfrac{1}{3}} \right)\left( \cot x-\sqrt{\dfrac{1}{3}} \right)=0\]
From the above equation, we can say that
$\cot x=-\sqrt{\dfrac{1}{3}}$ and $\cot x=\sqrt{\dfrac{1}{3}}$
We know that $\tan x=\dfrac{1}{\cot x}$. So the above equations can be written as
$\tan x=-\sqrt{3}$ and $\tan x=\sqrt{3}$
Also, the principle solutions of the equation $\tan x=\sqrt{3}$ is $x=\dfrac{\pi }{3}$.
From the first equation we have
$\Rightarrow \tan x=-\sqrt{3}$
We know that $\tan x$ is negative in the second and the fourth quadrants. So the solution of the above equation will lie in the second and the fourth quadrants, which can be respectively given by
$\begin{align}
  & x=\pi -\dfrac{\pi }{3},x=-\dfrac{\pi }{3} \\
 & \Rightarrow x=\dfrac{2\pi }{3},x=-\dfrac{\pi }{3} \\
\end{align}$
From the second equation, we have
$\tan x=\sqrt{3}$
We know that $\tan x$ is positive in the first and the third quadrants. So the solution of the above equation will lie in the first and the third quadrants, which can be respectively given by
\[\begin{align}
  & x=\dfrac{\pi }{3},x=\pi +\dfrac{\pi }{3} \\
 & \Rightarrow x=\dfrac{\pi }{3},x=\dfrac{4\pi }{3} \\
\end{align}\]

Hence, the solutions of the given equation are $x=\dfrac{2\pi }{3},x=-\dfrac{\pi }{3},x=\dfrac{\pi }{3},x=\dfrac{4\pi }{3}$.

Note: The interval given in the question is $0\le x\le 2\pi $. We know that there are infinite solutions possible corresponding to a trigonometric equation. So make sure that all the solutions lie in this interval only.