√(2)*sinX-√(2)*cosX=1

$\mathtt{\sqrt{2}sinx-\sqrt{2}cosx=1;~sinx-cosx=\frac{1}{\sqrt{2}};~}\\\\\mathtt{\sqrt{2}sin(x-\frac{\pi}{4})=\frac{\sqrt{2}}{2};~sin(x-\frac{\pi}{4})=\frac{1}{2};~\left[\begin{array}{ccc}\mathtt{x-\frac{\pi}{4}=\frac{\pi}{6}+2n\pi}\\\mathtt{x-\frac{\pi}{4}=\frac{2\pi}{3}+2n\pi}\end{array}\right}$

ОТВЕТ: $\mathtt{\left[\begin{array}{ccc}\mathtt{x=\frac{5\pi}{12}+2n\pi}\\\\\mathtt{x=\frac{11\pi}{12}+2n\pi}\end{array}\right,~n\in Z}$