Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 660

\[\boxed{\mathbf{660}\mathbf{.}}\]

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

\[y = \sin x:\]

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

\[y(2y + 1) = 0\]

\[y_{1} = 0\ \ и\ \ y_{2} = - \frac{1}{2}.\]

\[\sin x = 0\]

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

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

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

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

\[Ответ:\ \ \pi n;\ \ ( - 1)^{n + 1} \bullet \frac{\pi}{6} + \pi n.\]

\[2)\ 3\sin^{2}x - 5\sin x - 2 = 0\]

\[y = \sin x:\]

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

\[D = 25 + 24 = 49\]

\[y_{1} = \frac{5 - 7}{2 \bullet 3} = - \frac{2}{6} = - \frac{1}{3};\]

\[y_{2} = \frac{5 + 7}{2 \bullet 3} = 2.\]

\[1)\ \sin x = - \frac{1}{3}\]

\[x = ( - 1)^{n + 1} \bullet \arcsin\frac{1}{3} + \pi n.\]

\[2)\ \sin x = 2\]

\[корней\ нет.\]

\[Ответ:\ \ ( - 1)^{n + 1} \bullet \arcsin\frac{1}{3} + \pi n.\]

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

\[y = \cos x:\]

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

\[y(y - 2) = 0\]

\[y_{1} = 0\ \ и\ \ y_{2} = 2.\]

\[1)\ \cos x = 0\]

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

\[2)\ \cos x = 2\]

\[корней\ нет.\]

\[Ответ:\ \ \frac{\pi}{2} + \pi n.\]

\[4)\ 6\cos^{2}x + 7\cos x - 3 = 0\]

\[y = \cos x:\]

\[6y^{2} + 7y - 3 = 0\]

\[D = 49 + 72 = 121\]

\[y_{1} = \frac{- 7 - 11}{2 \bullet 6} = - \frac{18}{12} = - \frac{3}{2};\]

\[y_{2} = \frac{- 7 + 11}{2 \bullet 6} = \frac{4}{12} = \frac{1}{3}.\]

\[1)\ \cos x = - \frac{3}{2}\]

\[корней\ нет.\]

\[2)\ \cos x = \frac{1}{3}\]

\[x = \pm \arccos\frac{1}{3} + 2\pi n.\]

\[Ответ:\ \pm \arccos\frac{1}{3} + 2\pi n.\]