Step 1: Plan:
Square the equation after rearranging, then check which roots are real solutions.
Step 2: Solve:
$\cos\theta - 1 = \sin\theta$. Use half angles: $-2\sin^2\frac{\theta}{2} = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}$.
So $2\sin\frac{\theta}{2}\left(\sin\frac{\theta}{2}+\cos\frac{\theta}{2}\right) = 0$.
Either $\sin\frac{\theta}{2} = 0$, which gives $\theta = 2n\pi$, or $\tan\frac{\theta}{2} = -1$, which gives $\frac{\theta}{2} = n\pi - \frac{\pi}{4}$, i.e. $\theta = 2n\pi - \frac{\pi}{2}$.
Step 3: Check:
At $\theta = 2n\pi$: $1 - 0 = 1$. At $\theta = 2n\pi-\frac{\pi}{2}$: $0-(-1) = 1$. Both work, so the answer is option A.
Final Answer:
The general solution is $\theta = 2n\pi$ or $2n\pi - \frac{\pi}{2}$, option (A).
\[ \boxed{\theta = 2n\pi-\frac{\pi}{2}\ \text{or}\ \theta = 2n\pi} \]