ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1600

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\[1)\sin x + \cos x = \sqrt{1 + tg\ x}\]

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

\[\left( \sin x + \cos x \right)^{2} = \frac{\cos x}{\cos x} + \frac{\sin x}{\cos x}\]

\[1)\ \sin x + \cos x = 0\ \ \ \ \ |\ :\cos x\]

\[tg\ x + 1 = 0\]

\[tg\ x = - 1\]

\[x = - arctg\ 1 + \pi n\]

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

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

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

\[\sin x \bullet \cos x - \sin^{2}x = 0\]

\[\sin x \bullet \left( \cos x - \sin x \right) = 0.\]

\[Первое\ значение:\]

\[\sin x = 0\]

\[x = \arcsin 0 + \pi n = \pi n.\]

\[Второе\ значение:\]

\[\cos x - \sin x = 0\ \ \ \ \ |\ :\cos x\]

\[1 - tg\ x = 0\]

\[tg\ x = 1\]

\[x = arctg\ 1 + \pi n = \frac{\pi}{4} + \pi n.\]

\[Имеет\ смысл\ при:\]

\[\cos x \neq 0\]

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

\[1 + tg\ x \geq 0\]

\[tg\ x \geq - 1\]

\[\text{arctg\ }( - 1) + \pi n \leq x < \frac{\pi}{2} + \pi n\]

\[- \frac{\pi}{4} + \pi n \leq x < \frac{\pi}{2} + \pi n.\]

\[Имеет\ решения\ при:\]

\[\sin x + \cos x \geq 0\]

\[\sin x + \cos x = 0\ \ \ \ \ |\ :\cos x\]

\[tg\ x + 1 = 0\]

\[tg\ x = - 1\]

\[x = - arctg\ 1 + \pi n = - \frac{\pi}{4} + \pi n;\]

\[- \frac{\pi}{4} + 2\pi n \leq x \leq \frac{3\pi}{4} + 2\pi n.\]

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

\[2)\ \sqrt{5\sin{2x} - 2} = \sin x - \cos x\]

\[5\sin{2x} - 2 = \left( \sin x - \cos x \right)^{2}\]

\[5\sin{2x} - 2 =\]

\[= \sin^{2}x + \cos^{2}x - 2\sin x \bullet \cos x\]

\[5\sin{2x} - 2 = 1 - \sin{2x}\]

\[6\sin{2x} = 3\]

\[\sin{2x} = \frac{1}{2}\]

\[2x = ( - 1)^{n} \bullet \arcsin\frac{1}{2} + \pi n\]

\[2x = ( - 1)^{n} \bullet \frac{\pi}{6} + \pi n\]

\[x = \frac{1}{2} \bullet \left( ( - 1)^{n} \bullet \frac{\pi}{6} + \pi n \right)\]

\[x = ( - 1)^{n} \bullet \frac{\pi}{12} + \frac{\text{πn}}{2}\]

\[x_{1} = \frac{\pi}{12} + \frac{\pi}{2} \bullet (2n) = \frac{\pi}{12} + \pi n;\]

\[x_{2} = - \frac{\pi}{12} + \frac{\pi}{2} \bullet (2n + 1) =\]

\[= - \frac{\pi}{12} + \pi n + \frac{\pi}{2} = \frac{5\pi}{12} + \pi n.\]

\[Имеет\ смысл\ при:\]

\[5\sin{2x} - 2 \geq 0\]

\[\sin{2x} \geq \frac{2}{5}\]

\[\arcsin\frac{2}{5} + 2\pi n \leq 2x \leq \pi - \arcsin\frac{2}{5} + 2\pi n\]

\[\frac{\pi}{7} + 2\pi n \leq 2x \leq \pi - \frac{\pi}{7} + 2\pi n\]

\[\frac{\pi}{14} + \pi n \leq x \leq \frac{3\pi}{7} + \pi n.\]

\[Имеет\ решения\ при:\]

\[\sin x - \cos x \geq 0\]

\[\sin x - \cos x = 0\ \ \ \ \ |\ :\cos x\]

\[tg\ x - 1 = 0\]

\[tg\ x = 1\]

\[x = arctg\ 1 + \pi n = \frac{\pi}{4} + \pi n;\]

\[\frac{\pi}{4} + 2\pi n \leq x \leq \frac{5\pi}{4} + 2\pi n.\]

\[Ответ:\ \ \]

\[\frac{5\pi}{12} + 2\pi n;\ \ \frac{\pi}{12} + (2n + 1)\text{π.}\]