Using the addition formulae, \[\begin{aligned}\cos(2\theta-45^\circ)+\sin(2\theta-45^\circ)&=\frac{\cos2\theta+\sin2\theta}{\sqrt2}+\frac{\sin2\theta-\cos2\theta}{\sqrt2}\\&=\sqrt2\sin2\theta.\end{aligned}\] Thus \(\sin2\theta=-1/\sqrt2\). In \(-360^\circ\le2\theta\le360^\circ\), \[2\theta=-135^\circ,-45^\circ,225^\circ,315^\circ.\] Therefore \(\boxed{\theta=-67.5^\circ,-22.5^\circ,112.5^\circ,157.5^\circ}\).