How do you solve the following equation \[\cot \left( {\dfrac{x}{2}} \right) = 1\] in the interval \[[0,2\pi ] \] ?

Hint: The general solution of the trigonometric equation \[\tan x = \tan y\] is \[x = n\pi + y\] where \[n \in \mathbb{Z}\] . Arrange the given equation in the above form and use the result to get the general solution.

Complete step-by-step answer:
The equation contains the trigonometric function \[\cot x\] . So, we can here try to use the known general solution for the equation \[\tan x = \tan y\] and the result \[\cot x = \dfrac{1}{{\tan x}}\] . We will first try to write the equation in this form.
We know that \[\cot x = \dfrac{1}{{\tan x}}\] . So, the given equation becomes, \[\dfrac{1}{{\tan \left( {\dfrac{x}{2}} \right)}} = 1\] .
 \[ \Rightarrow \tan \left( {\dfrac{x}{2}} \right) = 1\]
We also know that \[\tan \left( {\dfrac{\pi }{4}} \right) = 1\] .
 \[ \Rightarrow \tan \left( {\dfrac{x}{2}} \right) = \tan \left( {\dfrac{\pi }{4}} \right)\]
Now, the equation is of the form \[\tan x = \tan y\] and we know that the solution of such an equation is \[x = n\pi + y\] where \[n \in \mathbb{Z}\] .
 \[ \Rightarrow \dfrac{x}{2} = n\pi + \dfrac{\pi }{4}\]
 \[ \Rightarrow x = 2n\pi + \dfrac{\pi }{2}\] which is the general solution of the given trigonometric equation.
But to find the solution that lies in the interval \[[0,2\pi ] \] we will look at the solutions generated by the general solution. Here we consider the solution generated by \[n = 0,1,2,...\] Because we are interested in finding the solution in the positive interval \[[0,2\pi ] \] .
The general solution gives us,
 \[x = 2(0)\pi + \dfrac{\pi }{2},{\text{ }}2(1)\pi + \dfrac{\pi }{2},{\text{ }}2(2)\pi + \dfrac{\pi }{2}...\] for \[n = 0,1,2,...\]
 \[ \Rightarrow x = \dfrac{\pi }{2},{\text{ }}2\pi + \dfrac{\pi }{2},{\text{ 4}}\pi + \dfrac{\pi }{2}...\]
 \[ \Rightarrow x = \dfrac{\pi }{2},{\text{ }}\dfrac{{5\pi }}{2},{\text{ }}\dfrac{{9\pi }}{2}...\]
Here, \[x = \dfrac{\pi }{2} \in [0,2\pi ] \] . So, it is the only solution in the interval \[[0,2\pi ] \] for the equation \[\cot \left( {\dfrac{x}{2}} \right) = 1\] .
Additional information:
Other important general solutions to solve the trigonometric equations are,
 \[\sin x = \sin y \Rightarrow x = n\pi + {( - 1)^n}y,{\text{ where }}n \in \mathbb{Z}\]
 \[\cos x = \cos y \Rightarrow x = 2n\pi \pm y,{\text{ where }}n \in \mathbb{Z}\]

Note: It is very important to keep the interval in mind while solving the equation. Because that itself would make the steps simpler and can help in choosing the required solution.