b) $\sin x-\cos x=\sqrt{2}\left(\dfrac{\sqrt{2}}{2}\sin x-\dfrac{\sqrt{2}}{2}\cos x\right)$$=\sqrt{2}(\sin x\cos \dfrac{\pi}{4}-\cos x\sin\dfrac{\pi}{4})=\sqrt{2}\sin \left( x-\dfrac{\pi}{4}\right)$$\sin x-\cos x=1\Leftrightarrow \sqrt{2}\sin\left( x-\dfrac{\pi}{4}\right) =1$$\Leftrightarrow \sin \left( x-\dfrac{\pi}{4}\right) =\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}=\sin\dfrac{\pi}{4}$$\Leftrightarrow\left[\begin{matrix} x-\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi \\ \\ x-\dfrac{\pi}{4}=\pi -\dfrac{\pi}{4}+k2\pi\end{matrix}\right.$$\Leftrightarrow\left[\begin{matrix}x= \dfrac{\pi}{2}+k2\pi \qquad (k\in Z) \\ \\ x=\pi +k2\pi\qquad (k\in Z)\end{matrix}\right.$
b) $\sin x-\cos x=\sqrt{2}\left(\dfrac{\sqrt{2}}{2}\sin x-\dfrac{\sqrt{2}}{2}\cos x\right)$$=\sqrt{2}(\sin x\cos \dfrac{\pi}{4}-\cos x\sin\dfrac{\pi}{4})=\sqrt{2}\sin \left( x-\dfrac{\pi}{4}\right)$$\sin x-\cos x=1\Leftrightarrow \sqrt{2}\sin\left( x-\dfrac{\pi}{4}\right) =1$$\Leftrightarrow \sin \left( x-\dfrac{\pi}{4}\right) =\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}=\sin\dfrac{\pi}{4}$$\Leftrightarrow\left[\begin{matrix} x-\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi \\ \\ x-\dfrac{\pi}{4}=\pi -\dfrac{\pi}{4}+k2\pi\end{matrix}\right.$$\Leftrightarrow\left[\begin{matriz}x= \dfrac{\pi}{2}+k2\pi \qquad (k\in Z) \\ \\ x=\pi +k2\pi\qquad (k\in Z)\end{matrix}\right.$
b) $\sin x-\cos x=\sqrt{2}\left(\dfrac{\sqrt{2}}{2}\sin x-\dfrac{\sqrt{2}}{2}\cos x\right)$$=\sqrt{2}(\sin x\cos \dfrac{\pi}{4}-\cos x\sin\dfrac{\pi}{4})=\sqrt{2}\sin \left( x-\dfrac{\pi}{4}\right)$$\sin x-\cos x=1\Leftrightarrow \sqrt{2}\sin\left( x-\dfrac{\pi}{4}\right) =1$$\Leftrightarrow \sin \left( x-\dfrac{\pi}{4}\right) =\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}=\sin\dfrac{\pi}{4}$$\Leftrightarrow\left[\begin{matrix} x-\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi \\ \\ x-\dfrac{\pi}{4}=\pi -\dfrac{\pi}{4}+k2\pi\end{matrix}\right.$$\Leftrightarrow\left[\begin{matri
x}x= \dfrac{\pi}{2}+k2\pi \qquad (k\in Z) \\ \\ x=\pi +k2\pi\qquad (k\in Z)\end{matrix}\right.$