$a)sin(x-60^o)=\dfrac{1}{2}=sin30^o$
$⇔$\(\left[ \begin{array}{l}x-60^o=30^o+k360^o\\x-60^o=180^o-30^o+k360^o\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=90^o+k360^o\\x=210^o+k360^o\end{array} \right.\) $(k∈Z)$
$b)cos(x-\dfrac{\pi}{4})=-\dfrac{1}{2}=cos(\dfrac{2\pi}{3})$
$⇔$\(\left[ \begin{array}{l}x-\dfrac{\pi}{4}=\dfrac{2\pi}{3}+k2\pi\\x-\dfrac{\pi}{4}=-\dfrac{2\pi}{3}+k2\pi\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=\dfrac{11\pi}{12}+k2\pi\\x=-\dfrac{5\pi}{12}+k2\pi\end{array} \right.\) $(k∈Z)$
$c)cos(x-2)=\dfrac{2}{5}$
$⇔$\(\left[ \begin{array}{l}x-2=arccos(\dfrac{2}{5})+k2\pi\\x-2=-arccos(\dfrac{2}{5})+k2\pi\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=2+arccos(\dfrac{2}{5})+k2\pi\\x=2-arccos(\dfrac{2}{5})+k2\pi\end{array} \right.\) $(k∈Z)$
