Giải: 1. `cot3x=\frac{-1}{\sqrt{3}}` 2. `cos3x=\frac{-\sqrt{3}}{2}`

1)

ĐK: $\sin3x\ne 0$

$\to x\ne \dfrac{k\pi}{3}$

$\cot3x=\dfrac{-1}{\sqrt3}$

$\to 3x=\dfrac{-\pi}{3}+k\pi$

$\to x=\dfrac{-\pi}{9}+\dfrac{k\pi}{3}$ (TM)

2)

$\cos3x=\dfrac{-\sqrt3}{2}$

$\to 3x=\pm\dfrac{5\pi}{6}+k2\pi$

$\to x=\pm\dfrac{5\pi}{18}+\dfrac{k2\pi}{3}$