`~rai~`
\(c)\cot\left(x+\dfrac{\pi}{4}\right)+\cot2x=0.\quad(1)\\ĐKXĐ:\begin{cases}\sin\left(x+\dfrac{\pi}{4}\right)\ne 0\\\sin 2x\ne 0\end{cases}\\\Leftrightarrow \begin{cases}x+\dfrac{\pi}{4}\ne k\pi\\2x\ne k\pi\end{cases}\\\Leftrightarrow \begin{cases}x\ne -\dfrac{\pi}{4}+k\pi\\x\ne k\dfrac{\pi}{2}.\end{cases}\quad(k\in\mathbb{Z})\\(1)\Leftrightarrow \cot\left(x+\dfrac{\pi}{4}\right)=-\cot2x\\\Leftrightarrow \cot\left(x+\dfrac{\pi}{4}\right)=\cot(-2x)\\\Leftrightarrow x+\dfrac{\pi}{4}=-2x+k\pi\\\Leftrightarrow 3x=-\dfrac{\pi}{4}+k\pi\\\Leftrightarrow x=-\dfrac{\pi}{12}+k\dfrac{\pi}{3}.\text{(thỏa mãn)}\quad\quad(k\in\mathbb{Z})\\\text{Vậy phương trình có họ nghiệm là:}x=-\dfrac{\pi}{12}+k\dfrac{\pi}{3}.(k\in\mathbb{Z})\)
