Solving $\sin(x)-\cos(x)=1$

You made a mistake:

$$ \sin x - \cos x=\sqrt2\sin\left(x\color{red}-\frac\pi4\right). $$

The correctness of the last expression can be easily verified by trigonometric summation formula:

$$ \sin(x+y)=\sin x \cos y+\cos x \sin y. $$

I would rather use the substitution $$\cos(x)=(\pm)\sqrt{1-\sin^2(x)}$$ to obtain $$\sin(x)-\cos(x)=1\iff\ldots\iff \sin(x)-1=(\pm)\sqrt{1-\sin^2(x)}$$ Squaring $$\sin^2(x)-2\sin(x)+1=1-\sin^2(x)\iff 2\sin^2(x)-2\sin(x)=0\iff \color{blue}{\sin^2(x)-\sin(x)=0}$$

Can you end it now?