$sin3x-cos3x=-1$
$⇔\sqrt[]{2}sin(3x-\dfrac{\pi}{4})=-1$
$⇔sin(3x-\dfrac{\pi}{4})=\dfrac{-1}{\sqrt[]{2}}=sin(-\dfrac{\pi}{4})$
$⇔$\(\left[ \begin{array}{l}3x-\dfrac{\pi}{4}=-\dfrac{\pi}{4}+k2\pi\\3x-\dfrac{\pi}{4}=\pi+\dfrac{\pi}{4}+k2\pi\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}3x=k2\pi\\3x=\dfrac{3\pi}{2}+k2\pi\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=\dfrac{k2\pi}{3}\\x=\dfrac{\pi}{2}+\dfrac{k2\pi}{3}\end{array} \right.\)
