Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 1265

\[\boxed{\mathbf{1265}\mathbf{.}}\]

\[1)\ 2\cos^{2}x + \sin x - 1 < 0\]

\[2\left( 1 - \sin^{2}x \right) + \sin x - 1 < 0\]

\[2 - 2\sin^{2}x + \sin x - 1 < 0\]

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

\[Пусть\ y = \sin x:\]

\[2y^{2} - y - 1 > 0\]

\[D = 1^{2} + 4 \bullet 2 = 1 + 8 = 9\]

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

\[\text{\ \ }y_{2} = \frac{1 + 3}{2 \bullet 2} = 1\]

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

\[y < - \frac{1}{2}\ или\ y > 1.\]

\[Первое\ неравенство:\]

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

\[- \pi - \arcsin\left( - \frac{1}{2} \right) +\]

\[+ 2\pi n < x < \arcsin\left( - \frac{1}{2} \right) + 2\pi n\]

\[- \pi + \arcsin\frac{1}{2} +\]

\[+ 2\pi n < x < - \arcsin\frac{1}{2} + 2\pi n\]

\[- \pi + \frac{\pi}{6} + 2\pi n < x < - \frac{\pi}{6} + 2\pi n\]

\[- \frac{5\pi}{6} + 2\pi n < x < - \frac{\pi}{6} + 2\pi n.\]

\[Второе\ неравенство:\]

\[\sin x > 1 - корней\ нет.\]

\[Ответ:\ - \frac{5\pi}{6} + 2\pi n < x < - \frac{\pi}{6} +\]

\[+ 2\pi n.\]

\[2)\ 2\sin^{2}x - 5\cos x + 1 > 0\]

\[2\left( 1 - \cos^{2}x \right) - 5\cos x + 1 > 0\]

\[2 - 2\cos^{2}x - 5\cos x + 1 > 0\]

\[2\cos^{2}x + 5\cos x - 3 < 0\]

\[Пусть\ y = \cos x:\]

\[2y^{2} + 5y - 3 < 0\]

\[D = 5^{2} + 4 \bullet 2 \bullet 3 = 25 + 24 =\]

\[= 49\]

\[y_{1} = \frac{- 5 - 7}{2 \bullet 2} = - 3\ \ и\ \ \]

\[y_{2} = \frac{- 5 + 7}{2 \bullet 2} = \frac{2}{4} = \frac{1}{2}.\]

\[(y + 3)\left( y - \frac{1}{2} \right) < 0\]

\[- 3 < y < \frac{1}{2}.\]

\[Первое\ неравенство:\]

\[\cos x > - 3 - при\ любом\ \text{x.}\]

\[Второе\ неравенство:\]

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

\[\arccos\frac{1}{2} + 2\pi n < x < 2\pi -\]

\[- \arccos\frac{1}{2} + 2\pi n\]

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

\[\frac{\pi}{3} + 2\pi n < x < \frac{5\pi}{3} + 2\pi n.\]

\[Ответ:\ \ \frac{\pi}{3} + 2\pi n < x < \frac{5\pi}{3} +\]

\[+ 2\pi n.\]