Đk tự làm nhé
$\cot x+\sin x=\dfrac{\cos x}{1-\cos x}+\dfrac{1}{\sin x}$
$\Leftrightarrow \cot x = \dfrac{\cos x}{1-\cos x}+\dfrac{1}{\sin x} -\sin x$
$\Leftrightarrow \cot x = \dfrac{\cos x}{1-\cos x}+ \dfrac{1-\sin^2 x}{\sin x}$
$\Leftrightarrow \cos x \bigg (\dfrac{1}{\sin x}- \dfrac{1}{1-\cos x} -\tan x \bigg )=0$