<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] $$
\begin{aligned}
\sin x-\cos x &=1 \\
-\frac{1}{\sqrt{2}} \sin x+\frac{1}{\sqrt{2}} \cos x &=-\frac{1}{\sqrt{2}} \\
\cos \left(x+\frac{\pi}{4}\right) &=-\frac{1}{\sqrt{2}} \\
x+\frac{\pi}{4} &=2 n \pi \pm \frac{3 \pi}{4} \\
x &=\frac{\pi}{2}(4 n+1) \text { or }(2 n-1) \pi, \text{ where }n\in Z.
\end{aligned}
$$
</s>