TITLE: Solve trigonometric equation $ \cot x + \cos x = 1 + \cot x \cos x $
QUESTION [5 upvotes]: 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).

REPLY [2 votes]: $$ \cot x + \cos x = 1 + \cot x \cos x $$

$$ \cot x + \cos x - 1 - \cot x \cos x =0$$
$$-( \cos x -1) (\cot x - 1) =0 $$
$$( \cos x -1) (\cot x - 1) =0 $$
Can you finish from here?

REPLY [2 votes]: 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