$a)sinx=-\dfrac{\sqrt[]{3}}{2}=sin(-\dfrac{\pi}{3})$
$⇔$\(\left[ \begin{array}{l}x=-\dfrac{\pi}{3}+k2\pi\\x=\pi+\dfrac{\pi}{3}+k2\pi\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=-\dfrac{\pi}{3}+k2\pi\\x=\dfrac{4\pi}{3}+k2\pi\end{array} \right.\) $(k∈Z)$
$b)cos(3x-\dfrac{\pi}{3})=-1$
$⇔3x-\dfrac{\pi}{3}=\pi+k2\pi$
$⇔3x=\dfrac{4\pi}{3}+k2\pi$
$⇔x=\dfrac{4\pi}{9}+\dfrac{k2\pi}{3}(k∈Z)$
$d)cot(x+\dfrac{\pi}{3}=\dfrac{\sqrt[]{3}}{3}=cot(\dfrac{\pi}{3})$
$⇔x+\dfrac{\pi}{3}=\dfrac{\pi}{3}+k\pi$
$⇔x=k\pi(k∈Z)$
$e)sinx=\dfrac{1}{4}$
$⇔$\(\left[ \begin{array}{l}x=arcsin(\dfrac{\pi}{4})+k2\pi\\x=\pi-arcsin(\dfrac{\pi}{4})+k2\pi\end{array} \right.\) $(k∈Z)$
$b)cos(x-\dfrac{\pi}{4})=-\dfrac{1}{2}=cos(\dfrac{2\pi}{3})$
$⇔$\(\left[ \begin{array}{l}x-\dfrac{\pi}{4}=\dfrac{2\pi}{3}+k2\pi\\x-\dfrac{\pi}{4}=-\dfrac{2\pi}{3}+k2\pi\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=\dfrac{11\pi}{12}+k2\pi\\x=-\dfrac{5\pi}{12}+k2\pi\end{array} \right.\) $(k∈Z)$
