$a)cos(x-1)=\dfrac{2}{3}$
$⇔$\(\left[ \begin{array}{l}x-1=arccos(\dfrac{2}{3})+k2\pi\\x-1=-arccos(\dfrac{2}{3})+k2\pi\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=1+arccos(\dfrac{2}{3})+k2\pi\\x=1+-arccos(\dfrac{2}{3})+k2\pi\end{array} \right.\)
$b)cos3x=cos12^o$
$⇔$\(\left[ \begin{array}{l}3x=12^o+k360^o\\3x=-12^o+k360^o\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=4^o+k120^o\\x=-4^o+k120^o\end{array} \right.\)
$c)cos^22x=\dfrac{1}{4}$
$cos2x=cos(\dfrac{\pi}{3})$ hay $cos2x=cos(\dfrac{2\pi}{3})$
$⇔$\(\left[ \begin{array}{l}2x=±\dfrac{\pi}{3}+k2\pi\\2x=±\dfrac{2\pi}{3}+k2\pi\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=±\dfrac{\pi}{6}+k\pi\\2x=±\dfrac{\pi}{3}+k\pi\end{array} \right.\)
