Answer
$(2k+1)\pi$, $\frac{\pi}{2}+2k\pi$
Work Step by Step
$\sin \theta-1=\cos \theta$
$\sin (2*\frac{\theta}{2})-1=\cos (2*\frac{\theta}{2})$
$2\sin \frac{\theta}{2}\cos \frac{\theta}{2}-1=2\cos^2\frac{\theta}{2}-1$
$2\sin \frac{\theta}{2}\cos \frac{\theta}{2}-2\cos^2\frac{\theta}{2}=0$
$2\cos\frac{\theta}{2}(\sin \frac{\theta}{2}-\cos\frac{\theta}{2})=0$
If $2\cos\frac{\theta}{2}=0$, then $\cos\frac{\theta}{2}=0$. So $\frac{\theta}{2}=\frac{\pi}{2}+k\pi$, which means that $\theta=\pi+2k\pi=(2k+1)\pi$.
If $\sin \frac{\theta}{2}-\cos\frac{\theta}{2}=0$, then $\sin \frac{\theta}{2}=\cos\frac{\theta}{2}$. So $\frac{\theta}{2}=\frac{\pi}{4}+k\pi$, which means $\theta=\frac{\pi}{2}+2k\pi$.
