Question

√2/2sinx-√2/2cosx=1решите уравнение.

Answer 1

$\frac{\sqrt2}{2}\cdot\sin x-\frac{\sqrt2}{2}\cdot\cos x=-1;\\ (\frac{\sqrt2}{2})^2+(\frac{\sqrt2}{2})^2=\frac{2}{4}+\frac24=\frac44=1;\\ \sin\frac\pi4=\cos\frac\pi4=\frac{\sqrt2}{2};\\ 1) \sin\frac\pi4\sin x-\cos\frac\pi4\cos x=-1;\\ -\cos(x+\frac\pi4)=-1;\\ \cos(x+\frac\pi4)=1;\\ x+\frac\pi4=2\pin;\\ x=-\frac\pi4+2\pi n, n\in Z;\\ 2)\cos\frac\pi4\sin x-\cos x\sin\frac\pi4=-1;\\ \sin(x-\frac\pi4)=-1;\\ x-\frac\pi4=-\frac\pi2+2pi n, n\in Z;\\ x=-\frac\pi2+\frac\pi4+2\pi n, n\in Z;$
$x=-\frac\pi4+2\pi n, n\in Z$

ответ:
$x=-\frac\pi4+2\pi n, n\in Z$