<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] Using your equation $$\sin x(\sin x + \cos x) = 1$$ the additional solutions you missed were $$\sin x = 1 \rightarrow x = \dfrac {\pi}{2}, \dfrac {3 \pi}{2}$$  The other solutions are correct.
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