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

When you square both sides of an equation, you may introduce extraneous solutions. Therefore, you must check that your solutions satisfy the original equations (a good idea in any case).

When you squared the equation $\sin\theta - 1 = \cos\theta$, you discovered that the resulting equation was satisfied when $\sin\theta = 0$ or $\sin\theta = 1$.

In the interval $[0, 2\pi)$, the equation $\sin\theta = 0$ is satisfied when $\theta = 0$ or $\theta = \pi$. If $\theta = 0$, then

$$\sin(0) - 1 = 0 - 1 = -1 \neq 1 = \cos(0)$$

so $0$ is an extraneous solution. On the other hand, if $\theta = \pi$, then

$$\sin(\pi) - 1 = 0 - 1 = -1 = \cos(\pi)$$

so $\theta = \pi$ is a valid solution.

In the interval $[0, 2\pi)$, the equation $\sin\theta = 1$ is satisfied when $\theta = \dfrac{\pi}{2}$. When $\theta = \dfrac{\pi}{2}$,

$$\sin\left(\frac{\pi}{2}\right) - 1 = 1 - 1 = 0 = \cos\left(\frac{\pi}{2}\right)$$

Thus, $\theta = \dfrac{\pi}{2}$ is a valid solution.

Thus, the general solution of the equation $\sin\theta - 1 = \cos\theta$ is

$$ \theta = \begin{cases} \pi + 2n\pi, n \in \mathbb{Z}\\ \dfrac{\pi}{2} + 2n\pi, n \in \mathbb{Z} \end{cases} $$

HINT:

by squaring both sides you are getting a solution set which is bigger than original problem

for example if $x^2=1$ both $x=-1$ and $1$ are solutions but you only want one of them

$$\begin{align} \sin(\theta)-\cos(\theta)&=1\\ \sin(\theta)\frac{1}{\sqrt{2}}-\cos(\theta)\frac{1}{\sqrt{2}}&=\frac{1}{\sqrt{2}}\\ \sin(\theta)\sin(\pi/4)-\cos(\theta)\cos(\pi/4)&=\frac{1}{\sqrt{2}}\\ -\cos(\theta+\pi/4)&=\frac{1}{\sqrt{2}}\\ \cos(\theta+\pi/4)&=-{\frac{1}{\sqrt{2}}}\\ \theta+\pi/4&=(2k+1)\pi\pm\frac{\pi}{4}\\ \theta&=(2k+3/4)\pi\pm\frac{\pi}{4}\\ \theta&=(2k+1/2)\pi\qquad\text{or}\qquad(2k+1)\pi \end{align}$$

Thinking of the unit circle as a compass, these solutions are "north" and "west".