How do you solve \[\csc x + \cot x = 1\] and find all solutions in the interval \[\left[ {0,2\pi } \right)\]?

Hint: Here, we will first convert the given trigonometric function into sine and cosine functions. We will then use basic mathematical formulas and suitable algebraic identity to convert the equation into a quadratic equation. Then we will solve the quadratic equation to find the required solution of the given equation.

Formula Used:
We will use the following formula:
1. Trigonometric Ratio: \[\cot x = \dfrac{{\cos x}}{{\sin x}}\]
2. Trigonometric Co-ratio: \[\csc x = \dfrac{1}{{\sin x}}\]
3. Trigonometric identity: \[{\sin ^2}x + {\cos ^2}x = 1\]

Complete Step by Step Solution:
We are given an equation \[\csc x + \cot x = 1\].
We know that the Trigonometric Ratio: \[\cot x = \dfrac{{\cos x}}{{\sin x}}\] and Trigonometric Co-ratio: \[\csc x = \dfrac{1}{{\sin x}}\]
Now, by rewriting the equation in terms of sine and cosine using the Trigonometric ratio and Trigonometric Co-ratio, we get
\[ \Rightarrow \dfrac{1}{{\sin x}} + \dfrac{{\cos x}}{{\sin x}} = 1\]
By adding the numerators in the fraction, we get
\[ \Rightarrow \dfrac{{1 + \cos x}}{{\sin x}} = 1\]
By rewriting the equation, we get
\[ \Rightarrow 1 + \cos x = \sin x\]
By squaring on both the sides of the equation, we get
\[ \Rightarrow {\left( {1 + \cos x} \right)^2} = {\left( {\sin x} \right)^2}\]
The square of the sum of the numbers is given by an algebraic identity \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
Now, by using the algebraic identity, we get
\[ \Rightarrow 1 + {\left( {\cos x} \right)^2} + 2\cos x = {\left( {\sin x} \right)^2}\]
\[ \Rightarrow 1 + {\cos ^2}x + 2\cos x = {\sin ^2}x\]
We know that Trigonometric identity: \[{\sin ^2}x + {\cos ^2}x = 1\]
\[ \Rightarrow {\sin ^2}x = 1 - {\cos ^2}x\]
Now, by using the Trigonometric identity, we get
\[1 + {\cos ^2}x + 2\cos x = 1 - {\cos ^2}x\]
By rewriting the equation, we get
\[ \Rightarrow 1 + {\cos ^2}x + 2\cos x - 1 + {\cos ^2}x = 0\]
Adding and subtracting the like terms, we get
\[ \Rightarrow 2{\cos ^2}x + 2\cos x = 0\]
By taking out the common factors, we get
\[ \Rightarrow 2\cos x\left( {\cos x + 1} \right) = 0\]
By using the zero product property, we get
\[ \Rightarrow 2\cos x = 0\] or \[\cos x + 1 = 0\]
\[ \Rightarrow \cos x = 0\] or \[\cos x = - 1\]
\[ \Rightarrow x = {\cos ^{ - 1}}\left( 0 \right)\] or \[x = {\cos ^{ - 1}}\left( { - 1} \right)\]
We know that \[\cos n\dfrac{\pi }{2} = 0\] where\[n\] is odd and \[\cos \pi = - 1\] , we get
\[ \Rightarrow x = n\dfrac{\pi }{2}\] or \[x = \pi \]
Since it is given that the solutions of the equations should lie in the interval \[\left[ {0,\left. {2\pi } \right)} \right.\], so we get
\[ \Rightarrow x = \dfrac{\pi }{2},\dfrac{{3\pi }}{2}\] or \[x = \pi \] where \[n = 1,3\] and \[x \in \left[ {0,2\pi } \right)\]
When \[x = \dfrac{{3\pi }}{2}\] and \[x = \pi \], then the Trigonometric equation \[\csc x + \cot x = 1\]is not defined.
The only solution satisfying the given Trigonometric equation \[\csc x + \cot x = 1\] is \[x = \dfrac{\pi }{2}\] where \[x \in \left[ {0,2\pi } \right)\].

Therefore, the solutions of the equation \[\csc x + \cot x = 1\] in the interval \[\left[ {0,\left. {2\pi } \right)} \right.\] is \[\dfrac{\pi }{2}\].

Note:
We know that Trigonometric Equation is defined as an equation involving trigonometric ratios. Trigonometric identity is an equation that is always true for all the variables. Trigonometric Ratios of a Particular angle are the ratios of the sides of a right-angled triangle with respect to any of its acute angle. We know that the inverse trigonometric function is used to find the missing angles in a right-angled triangle. We should know that an Open interval is an interval that does not include the endpoints and is denoted by () whereas a Closed interval is an interval that includes the endpoints and is denoted by [].Zero product property states that when the product of two factors is zero, then one of the factor is separately zero i.e., if \[ab = 0\] then \[a = 0\] or \[b = 0\].