Solving $\cos x+\sin x-1=0$

Rewrite as:

$$\cos(x)+\sin(x)=1$$ Then square: $$\cos^2(x)+2\cos(x)\sin(x)+\sin^2(x)=1$$ Note an important identity: $$1+2\cos(x)\sin(x)=1$$ Then simplify and note another identity: $$\sin(2x)=0$$

Can you take it from here?

Given $$\color{blue}{\cos x+\sin x-1=0} $$$$\cos x+\sin x=1 $$ Divide both sides by $\color{blue}{\sqrt{2}}$ we get $$\frac{1}{\sqrt{2}}\cos x+ \frac{1}{\sqrt{2}}\sin x=\frac{1}{\sqrt{2}}$$ $$\cos x\cos\frac{\pi}{4}+\sin x\sin\frac{\pi}{4}=\cos\frac{\pi}{4}$$ Using formula $\color{purple}{\cos A\cos B+\sin A\sin B=\cos(A-B)}$, we get $$\color{green}{\cos\left(x-\frac{\pi}{4}\right)=\cos\frac{\pi}{4}}$$ As there is no information about the unknown value $x$ hence writing the general solutions as follows $$x-\frac{\pi}{4}=2n\pi\pm \frac{\pi}{4}$$$$x=2n\pi\pm \frac{\pi}{4}+\frac{\pi}{4}$$$$ \color{}{x=2n\pi} \quad \text{Or}\quad \color{}{x=2n\pi+\frac{\pi}{2}} $$ $$\color{blue}{x\in\{2n\pi\}\cup\{2n\pi+\frac{\pi}{2}\}}$$ Where, $\color{}{n \space \text{is any integer}}$ i.e. $\ n=0, \pm1, \pm2,\pm3, \ldots$

As $\sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\left(\cos x+\sin x\right)$ by the angle addition formula we find that: \begin{equation} \begin{aligned} \cos x+\sin x-1&=0\\ \implies\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)&=1\\ \implies\sin\left(x+\frac{\pi}{4}\right)&=\frac{\sqrt{2}}{2}\\ \implies x+\frac{\pi}{4}&=\frac{\pi}{4}+2\pi n,\frac{3\pi}{4}+2\pi n\\ \implies x&=2\pi n,\frac{\pi}{2}+2\pi n \end{aligned} \end{equation} for $n\in\mathbb{Z}$.