<s>[INST] $( 1 - \sin x ) ( 1 + \csc x ) = \cos x \; , \; 0 \leq x \leq 2 \pi$ I found some solutions to this problem however I appear to be missing some solutions. Could anyone help me find the whole solution set?
$$( 1 - \sin x ) ( 1 + \csc x ) = \cos x \; , \; 0 \leq x \leq 2 \pi$$
$$1 + \csc x - \sin x - \sin x \csc x = \cos x $$
$$ 1 + \frac {1}{\sin x} - \sin x - 1 = \cos x $$
$$ \frac {1}{\sin x} - \sin x = \cos x $$
$$ \frac {1}{\sin x} = \cos x + \sin x $$
$$ 1 = \sin x (\cos x + \sin x) $$
$$ \sin^2 x + \cos^2 x = \sin x \cos x + \sin^2 x $$
$$ \cos^2 x = \sin x \cos x $$
$$ \cos x = \sin x $$
$$ \frac {\sin x}{\cos x} = 1 $$
$$\tan x = 1$$
$$ x = \frac \pi 4, \, \frac {5 \pi}{4}$$
 [/INST] *

*

$$( 1 - \sin x ) ( 1 + \csc x ) = \cos x \; , \; 0 \leq x \leq 2 \pi\tag{*}\\\ldots$$
$$ \cos^2 x = \sin x \cos x \tag A$$
$$ \cos x = \sin x $$
$$ \frac {\sin x}{\cos x} = 1 \tag B$$
$$\tan x = 1$$
$$ x = \frac \pi 4, \, \frac {5 \pi}{4}$$

You divided equation $(A)$ by $\cos^2x$ to obtain equation $(B).$ Notice that this move implicitly assumes that the former has no solution satisfying $\cos^2x=0.$ But this assumption is unjustifiable.
You ought to have factorised it out instead:
$$ \cos^2 x = \sin x \cos x \tag A$$
$$ \cos^2 x \big(\ 1-\tan x \big)=0\\
   \cos x=0 \;\text{or}\;\tan x =1$$
$$x=\frac \pi 4\:\text{or}\:\frac \pi 2\:\text{or}\:\frac {5 \pi}4\:\text{or}\:\frac {3 \pi}2\tag#$$


*But we are not quite done. Notice that our work so far has the following structure
   if Line $1$ is true, then so is Line $2$
   if Line $2$ is true, then so is Line $3$
   ...
   if Line $n{-}1$ is true, then so is Line $n$
but nothing indicates that Line $n$ (the solution statement) actually implies Line $1$ (the given equation)?
This means that we have potentially introduced extraneous solutions. Thus, to finish our work, we must check each solution in $(\#)$ to see whether it actually satisfies the given equation $(*).$ In this example, all the four solutions happen to be genuine.
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