$d)cos(2x+\dfrac{\pi}{3})=-\dfrac{1}{2}=cos(\dfrac{2\pi}{3})$
$⇔$\(\left[ \begin{array}{l}2x+\dfrac{\pi}{3}=\dfrac{2\pi}{3}+k2\pi\\2x+\dfrac{\pi}{3}=-\dfrac{2\pi}{3}+k2\pi\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}2x=\dfrac{\pi}{3}+k2\pi\\2x=-\pi+k2\pi\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=\dfrac{\pi}{6}+k\pi\\x=-\dfrac{\pi}{2}+k\pi\end{array} \right.\) $(k∈Z)$
$e)cos(2x+50^o)=\dfrac{1}{2}=cos60^o$
$⇔2x+50^o=±60^o+k360^o$
$⇔$\(\left[ \begin{array}{l}2x=10^o+k360^o\\2x=-110^o+k360^o\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=5^o+k120^o\\x=-55^o+k120^o\end{array} \right.\) $(k∈Z)$
$i)cos(x-\dfrac{\pi}{4})=sin(2x+\dfrac{\pi}{2})$
$cos(x-\dfrac{\pi}{4})=cos(-2x)$
$⇔$\(\left[ \begin{array}{l}x-\dfrac{\pi}{4}=-2x+k2\pi\\x-\dfrac{\pi}{4}=2x+k2\pi\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}3x=\dfrac{\pi}{4}+k2\pi\\-x=\dfrac{\pi}{4}+k2\pi\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=\dfrac{\pi}{12}+\dfrac{k2\pi}{3}\\x=-\dfrac{\pi}{4}-k2\pi\end{array} \right.\) $(k∈Z)$
