Đáp án:
$\left[\begin{array}{l}x = \dfrac{\pi}{2} + k2\pi\\x= -\dfrac{3\pi}{4}+ k3\pi\end{array}\right.\quad (k\in\Bbb Z)$
Giải thích các bước giải:
$\left(\cot\dfrac{x}{2} - 1\right)\cdot\left(\cot\dfrac{x}{3} +1\right)=0\qquad (*)$
$ĐKXĐ:\, \begin{cases}\sin\dfrac{x}{2}\ne 0\\\sin\dfrac{x}{3}\ne 0\end{cases}\Leftrightarrow x \ne n\pi \quad (n\in \Bbb Z)$
$(*)\Leftrightarrow \left[\begin{array}{l}\cot\dfrac{x}{2} = 1\\\cot\dfrac{x}{3} = -1\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\dfrac{x}{2} = \dfrac{\pi}{4} + k\pi\\\dfrac{x}{3} = -\dfrac{\pi}{4}+ k\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{2} + k2\pi\\x= -\dfrac{3\pi}{4}+ k3\pi\end{array}\right.\quad (k\in\Bbb Z)$
