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Solve trigonometric equation: $$ \cot (x) + \cos (x) = 1 + \cot (x) \cos (x) $$ I tried to multiply both sides with $\sin x$ (which I'm not sure if I can multiply with sin).

3SAT
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Gjekaks
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3 Answers3

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$$(\cot x-1)(\cos x-1)=0$$

$\cos x=1=\cos0\implies x=2m\pi\pm0=2m\pi$ where $m$ is any integer

$\cot x=1\iff\tan x=1=\tan\dfrac\pi4\implies x=n\pi+\dfrac\pi4$ where $n$ is any integer

lab bhattacharjee
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  • Dear sir, how did you figure it out so fast that you can write it like this $(\cot x−1)(\cos x−1)$? – Gjekaks Aug 12 '15 at 07:44
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    @Gjekask practice, practice, practice...after some time you just "see" things like this. One possible way including some steps between: $$\cot(x)+\cos(x)=1+\cot(x)\cos(x)\Leftrightarrow a+b=1+ab \Leftrightarrow a+b-ab-1=0.$$ As we have $a$ and $b$ as well as $ab$ one might think of finding a factorization and applying a similar method as Vieta's formulas we can easily find $$a+b-ab-1=-(a-1)(b-1).$$ – Hirshy Aug 12 '15 at 07:53
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$$ \cot x + \cos x = 1 + \cot x \cos x $$

$$ \cot x + \cos x - 1 - \cot x \cos x =0$$

$$-( \cos x -1) (\cot x - 1) =0 $$

$$( \cos x -1) (\cot x - 1) =0 $$

Can you finish from here?

3SAT
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2

Notice, we have $$\cot x+\cos x=1+\cot x\cos x$$ $$\frac{\cos x}{\sin x}+\cos x=1+\frac{\cos x}{\sin x}\cos x $$ $$\cos x+\sin x\cos x=\sin x+\cos^2 x $$ $$\cos x+\sin x\cos x-\sin x-\cos^2 x=0 $$ $$\underbrace{\cos x-\sin x}+\underbrace{\sin x\cos x-\cos^2 x}=0 $$

$$(\cos x-\sin x)-\cos x(\cos x-\sin x)=0 $$ $$(1-\cos x)(\cos x-\sin x)=0 $$ $$1-\cos x=0\vee \cos x-\sin x=0$$ $$\cos x=1\vee \sin x=\cos x\iff \tan x=1$$ $$x=2k\pi\vee x=k\pi+\frac{\pi}{4}$$ For all, $k$ is any integer

Harish Chandra Rajpoot
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