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?
 A: 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 
A: 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 
$$
\theta = 
\begin{cases}
\pi + 2n\pi, n \in \mathbb{Z}\\
\dfrac{\pi}{2} + 2n\pi, n \in \mathbb{Z}
\end{cases}
$$
A: $$\begin{align}
\sin(\theta)-\cos(\theta)&=1\\
\sin(\theta)\frac{1}{\sqrt{2}}-\cos(\theta)\frac{1}{\sqrt{2}}&=\frac{1}{\sqrt{2}}\\
\sin(\theta)\sin(\pi/4)-\cos(\theta)\cos(\pi/4)&=\frac{1}{\sqrt{2}}\\
-\cos(\theta+\pi/4)&=\frac{1}{\sqrt{2}}\\
\cos(\theta+\pi/4)&=-{\frac{1}{\sqrt{2}}}\\
\theta+\pi/4&=(2k+1)\pi\pm\frac{\pi}{4}\\
\theta&=(2k+3/4)\pi\pm\frac{\pi}{4}\\
\theta&=(2k+1/2)\pi\qquad\text{or}\qquad(2k+1)\pi
\end{align}$$
Thinking of the unit circle as a compass, these solutions are "north" and "west".
A: Avoid squaring as reasoned by N. F. Taussig.
$$\sin\theta=1+\cos\theta$$
Using double angle formulae,
$$2\sin\dfrac\theta2\cos\dfrac\theta2-2\cos^2\dfrac\theta2=0$$
$$\iff2\cos\dfrac\theta2\left(\sin\dfrac\theta2-\cos\dfrac\theta2\right)=0$$
Now the product of two terms is zero,
If $\cos\dfrac\theta2=0,\dfrac\theta2=(2n+1)\dfrac\pi2\iff\theta=(2n+1)\pi$
Else $\sin\dfrac\theta2-\cos\dfrac\theta2=0\iff\sin\dfrac\theta2=\cos\dfrac\theta2\iff\tan\dfrac\theta2=1$
Now if $\tan x=\tan A,x=m\pi+A$
where $m,n$ are arbitrary integer
A: use that $$\sin(\theta)=2\,{\frac {\tan \left( \theta/2 \right) }{1+ \left( \tan \left( \theta
/2 \right)  \right) ^{2}}}
$$
and $$\cos(\theta)={\frac {1- \left( \tan \left( \theta/2 \right)  \right) ^{2}}{1+
 \left( \tan \left( \theta/2 \right)  \right) ^{2}}}
$$
