Answer
$2k\pi$, $\frac{3\pi}{2}+2k\pi$
Work Step by Step
$\cos \theta-\sin\theta=1$
$\cos(2*\frac{\theta}{2})-\sin(2*\frac{\theta}{2})=1$
$1-2\sin^2\frac{\theta}{2}-2\sin\frac{\theta}{2}\cos\frac{\theta}{2}=1$
$-2\sin^2\frac{\theta}{2}-2\sin\frac{\theta}{2}\cos\frac{\theta}{2}=0$
$2\sin^2\frac{\theta}{2}+2\sin\frac{\theta}{2}\cos\frac{\theta}{2}=0$
$2\sin\frac{\theta}{2}(\sin \frac{\theta}{2}+\cos \frac{\theta}{2})=0$
If $2\sin\frac{\theta}{2}=0$, then $\sin \frac{\theta}{2}=0$. So $\frac{\theta}{2}=k\pi$, which means $\theta=2k\pi$.
If $\sin\frac{\theta}{2}+\cos\frac{\theta}{2}=0$, then $\sin\frac{\theta}{2}=-\cos\frac{\theta}{2}$. This means that $\frac{\theta}{2}=\frac{3\pi}{4}+k\pi$, so $\theta=\frac{3\pi}{2}+2k\pi$.
