$cot(\dfrac{\pi}{2}-3x)=0$
Điều kiện $sinx\ne0$
$⇔x\ne k\pi$
$⇔\dfrac{\pi}{2}-3x=\dfrac{\pi}{2}+k\pi$
$⇔x=k3\pi(k∈Z)(l)$
$cot(x+2)=3$
Điều kiện $sin(x+2)\ne0$
$⇔x+2\ne k\pi$
$⇔x\ne-2+ k\pi$
$⇔x+2=arccot(3)+k\pi$
$⇔x=-2+arccot(3)+k\pi$
$cot(x-\dfrac{\pi}{6})=\dfrac{1}{\sqrt[]{3}}=cot(\dfrac{\pi}{3})$
Điều kiện $sin(x-\dfrac{\pi}{6})\ne0$
$⇔x-\dfrac{\pi}{6}\ne k\pi$
$⇔x\ne \dfrac{\pi}{6}+k\pi$
$⇔x-\dfrac{\pi}{6}=\dfrac{\pi}{3}+k\pi$
$⇔x=\dfrac{\pi}{2}+k\pi(k∈Z)$