ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 1227

\[\boxed{\mathbf{1227}\mathbf{.}}\]

\[1)\sin^{2}x + 2\sin x > 0\]

\[Пусть\ y = \sin x:\]

\[y^{2} + 2y > 0\]

\[y(y + 2) > 0\]

\[y < - 2\ или\ y > 0\]

\[Первое\ неравенство:\]

\[\sin x < - 2 - решений\ нет;\]

\[\sin x \neq - 2\]

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

\[Второе\ неравенство:\]

\[\sin x > 0\]

\[\arcsin 0 + 2\pi n < x < \pi -\]

\[- \arcsin 0 + 2\pi n\]

\[2\pi n < x < \pi + 2\pi n.\]

\[Ответ:\ \ 2\pi n < x < \pi + 2\pi n.\]

\[2)\cos^{2}x - \cos x \leq 0\]

\[Пусть\ y = \cos x:\]

\[y^{2} - y \leq 0\]

\[y(y - 1) \leq 0\]

\[0 \leq y \leq 1\]

\[Первое\ неравенство:\]

\[\cos x \geq 0\]

\[- \arccos 0 +\]

\[+ 2\pi n \leq x \leq \arccos a + 2\pi n\]

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

\[Второе\ неравенство:\]

\[\cos x \leq 1 - при\ любом\ x.\]

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

\[+ 2\pi n.\]

\[3)\ 2\sin^{2}x - \sin x - 3 < 0\]

\[Пусть\sin x = y:\]

\[2y^{2} - y - 3 < 0\]

\[D = 1 + 24 = 25\]

\[y_{1} = \frac{1 + 5}{4} = \frac{3}{2};\ \ \ \]

\[y_{2} = \frac{1 - 5}{4} = - 1.\]

\[(y + 1)(y - 1,5) < 0\]

\[- 1 < y < 1,5.\]

\[- 1 < \sin y < 1,5\]

\[\text{sin\ }x \neq - 1\]

\[x \neq - \pi + 2\pi n.\]

\[Ответ:x - любое\ число,\]

\[\ кроме\ \ x = - \pi + 2\pi n.\]

\[4)\ 2\cos^{2}x - 3\cos x - 2 > 0\]

\[Пусть\cos x = y:\]

\[2y^{2} - 3y - 2 > 0\]

\[D = 9 + 16 = 25\]

\[y_{1} = \frac{3 + 5}{4} = 2;\ \ \ \]

\[y_{2} = \frac{3 - 5}{4} = - \frac{1}{2}\]

\[\left( y + \frac{1}{2} \right)(y - 2) > 0\]

\[y < - \frac{1}{2};\ \ \ y > 2.\]

\[\cos x < - \frac{1}{2}\]

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

\[Ответ:\ \frac{2\pi}{3} + 2\pi n < x < \frac{4\pi}{3} +\]

\[+ 2\pi n.\]