$cos(x-2) =sin 3\bigg(x-\dfrac{\pi}{2}\bigg)$
$⇔cos(x-2) =sin \bigg(3x-\dfrac{3\pi}{2}\bigg)$
$⇔cos(x-2) =cos \bigg[\dfrac{\pi}{2}-(3x-\dfrac{3\pi}{2})\bigg]$
$⇔cos(x-2) =cos \bigg[\dfrac{\pi}{2}-(3x-\dfrac{3\pi}{2})\bigg]$
$⇔cos(x-2)=cos(2\pi-3x)$
$⇔$\(\left[ \begin{array}{l}x-2=2\pi-3x+k2\pi\\x-2=3x-2\pi+k2\pi\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}4x=2+2\pi+k2\pi\\-2x=2-2\pi+k2\pi\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=\dfrac{1}{2}+\dfrac{\pi}{2}+\dfrac{k\pi}{2}\\x=-1+\pi-k\pi\end{array} \right.\) $(k∈Z)$
