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

theonerishi
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    The step of dividing by $\cos x$ may have eliminated solutions, I would recommend revisiting the line prior to that step. – abiessu Jan 23 '22 at 16:13
  • @abiessu Thank you for your advice. I can't seem to figure out what to do instead. Can you provide another hint? – theonerishi Jan 23 '22 at 16:20
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    I'm not certain of the results yet, but the double-angle formulas for $\sin,\cos$ seem to be relevant in that case. – abiessu Jan 23 '22 at 16:26
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    Instead of dividing by $\cos x$, subtract $\sin x \cos x$ from both sides and then factor to get $\cos x(\cos x-\sin x)=0$. Then either $\cos x=0$ or $\cos x=\sin x$. In similar situations, rather than divide by a common factor on each side of the equation, instead, subtract, then divide by the common factor. – John Wayland Bales Jan 23 '22 at 17:28

2 Answers2

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

    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: $$ \cos^2 x = \sin x \cos x \tag A$$ $$ \cos^2 x \big(\ 1-\tan x \big)=0\\ \cos x=0 \;\text{or}\;\tan x =1$$ $$x=\frac \pi 4\:\text{or}\:\frac \pi 2\:\text{or}\:\frac {5 \pi}4\:\text{or}\:\frac {3 \pi}2\tag#$$

  2. 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.

ryang
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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.

bjcolby15
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