ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1374

\[\boxed{\mathbf{1374}\mathbf{.}}\]

\[\frac{\cos{2x}\ }{1 - \sin{2x}} = \cos x + \sin x\]

\[\cos{2x} = \left( \cos x + \sin x \right)\left( 1 - \sin{2x} \right)\]

\[\cos{2x} \bullet \left( \sin x - \cos x + 1 \right) = 0\]

\[\cos{2x} = 0\]

\[2x = \arccos 0 + \pi n = \frac{\pi}{2} + \pi n\]

\[x = \frac{1}{2} \bullet \left( \frac{\pi}{2} + \pi n \right) = \frac{\pi}{4} + \frac{\text{πn}}{2}.\]

\[\sin x - \cos x + 1 = 0\]

\[\cos x - \sin x = 1\ \ \ \ \ |\ :\sqrt{2}\]

\[\frac{\sqrt{2}}{2}\cos x - \frac{\sqrt{2}}{2}\sin x = \frac{\sqrt{2}}{2}\]

\[\cos\frac{\pi}{4} \bullet \cos x - \sin\frac{\pi}{4} \bullet \sin x = \frac{\sqrt{2}}{2}\]

\[\cos\left( \frac{\pi}{4} + x \right) = \frac{\sqrt{2}}{2}\]

\[\frac{\pi}{4} + x = \pm \arccos\frac{\sqrt{2}}{2} + 2\pi n =\]

\[= \pm \frac{\pi}{4} + 2\pi n;\]

\[x_{1} = - \frac{\pi}{4} + 2\pi n - \frac{\pi}{4} =\]

\[= - \frac{\pi}{2} + 2\pi n;\]

\[x_{2} = - \frac{\pi}{4} + 2\pi n + \frac{\pi}{4} = 2\pi n.\]

\[Имеет\ смысл\ при:\]

\[1 - \sin{2x} \neq 0\]

\[\sin{2x} \neq 1\]

\[2x \neq \arcsin 1 + 2\pi n\]

\[2x = \frac{\pi}{2} + 2\pi n\]

\[x \neq \frac{1}{2} \bullet \left( \frac{\pi}{2} + 2\pi n \right) \neq \frac{\pi}{4} + \pi n.\]

\[Ответ:\ \ \]

\[- \frac{\pi}{4} + \pi n;\ \ - \frac{\pi}{2} + 2\pi n;\ \ \ 2\pi n.\]