Question: 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

Answer: 
I hope you know the expansion$$\sin(\theta+\varphi)=\sin\theta\cos\varphi+\sin\varphi\cos\theta$$Because we can use it in the equation and greatly simplify the algebra needed. Rewriting your equation$$\sin x-\cos x=1\quad\implies\quad\sqrt2\left(\frac {\sin x}{\sqrt2}-\frac {\cos x}{\sqrt2}\right)=1$$The expression inside the parenthesis can be rewritten as$$\sin x\cos\frac {\pi}4-\cos x\sin\frac {\pi}4=\sin\left(x-\frac {\pi}4\right)$$Hence$$\sqrt2\sin\left(x-\frac {\pi}4\right)=1$$Can you take it from here?
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

$x = π/2 + 2πk.$