$1)cosx-sinx=\sqrt[]{2}cos3x$
$⇔\dfrac{1}{\sqrt[]{2}}cosx-\dfrac{1}{\sqrt[]{2}}sinx=cos3x$
$⇔cosx.cos(\dfrac{\pi}{4})-sinx.sin(\dfrac{\pi}{4})=cos3x$
$⇔cos(x-\dfrac{\pi}{4})=cos3x$
$⇔$\(\left[ \begin{array}{l}x-\dfrac{\pi}{4}=3x+k2\pi\\x-\dfrac{\pi}{4}=-3x+k2\pi\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}-2x=\dfrac{\pi}{4}+k2\pi\\4x=\dfrac{\pi}{4}+k2\pi\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=-\dfrac{\pi}{8}-k\pi\\x=\dfrac{\pi}{16}+\dfrac{k\pi}{2}\end{array} \right.\) $(k∈Z)$