<s>[INST] Solving $\sin x - \cos x = 1$ I have been trying to solve $\sin x - \cos x = 1$ by squaring both sides but has not been able to obtain the solution. Here is what I did:
$$\begin{align}(\sin x - \cos x)^2 &=1^2\\\sin^2x-2\sin x\cos x+\cos^2x&=1\\1-\sin 2x&=1\\\sin2x&=0\end{align}$$
Obviously $x=0$ is not a solution. May I ask why this is the case or where did things go wrong?
Thank You
 [/INST] Let $a:= \sin x -\cos x;$ $ b=1;$
You have $a = b.$
Squaring: $a^2 =b^2$ gives  
$(a-b)(a+b)=0. $
You find solutions for the squared equation:
$a=b$, or $a = -b.$
Need to be a bit careful:
$\sin 2x =0;$ or $x=0;$ implies:
1) $x = 2πk$, $k$ integer.
These are the solutions for $a = -b.$, I.e.
$\sin x - \cos x = -1$.
The solution to the original problem,
$\sin x - \cos x= +1:$
2) $2x = π$ ; or $x = π/2; $ or
$x = π/2 + 2πk.$
</s>