Solving $\sin\theta -1=\cos\theta $ 
Solve$$\sin\theta -1=\cos\theta $$

Steps I took to solve this:
$$\sin^{ 2 }\theta -2\sin\theta +1=1-\sin^2\theta $$
$$2\sin^{ 2 }\theta -2\sin\theta =0$$
$$(2\sin\theta )(\sin\theta -1)=0$$
$$2\sin\theta =0, \sin\theta -1=0$$ 
$$\quad \sin\theta =0, \sin\theta =1$$
$$\theta =0+\pi k, \theta =\frac { \pi  }{ 2 } +2\pi k$$
Why is $\theta =0+\pi k$ wrong?

$$\begin{align}
\sin(\theta)-\cos(\theta)&=1\\
\sin(\theta)\frac{1}{\sqrt{2}}-\cos(\theta)\frac{1}{\sqrt{2}}&=\frac{1}{\sqrt{2}}\\
\sin(\theta)\sin(\pi/4)-\cos(\theta)\cos(\pi/4)&=\frac{1}{\sqrt{2}}\\
-\cos(\theta+\pi/4)&=\frac{1}{\sqrt{2}}\\
\cos(\theta+\pi/4)&=-{\frac{1}{\sqrt{2}}}\\
\theta+\pi/4&=(2k+1)\pi\pm\frac{\pi}{4}\\
\theta&=(2k+3/4)\pi\pm\frac{\pi}{4}\\
\theta&=(2k+1/2)\pi\qquad\text{or}\qquad(2k+1)\pi
\end{align}$$
Thinking of the unit circle as a compass, these solutions are "north" and "west".
