ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1602

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

\[\frac{- 2\cos^{2}x + \cos x + 1}{\cos^{2}x} = 0\]

\[2\cos^{2}x - \cos x - 1 = 0\]

\[y = \cos x:\]

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

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

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

\[y_{2} = \frac{1 + 3}{2 \bullet 2} = \frac{4}{4} = 1.\]

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

\[x = \pm \left( \pi - \arccos\frac{1}{2} \right) + 2\pi n\]

\[x = \pm \left( \pi - \frac{\pi}{3} \right) + 2\pi n\]

\[x = \pm \frac{2\pi}{3} + 2\pi n;\]

\[x_{1} = - \frac{2\pi}{3} + 2\pi n =\]

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

\[= \frac{\pi}{3} + (2n - 1)\pi;\]

\[x_{2} = \frac{2\pi}{3} + 2\pi n =\]

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

\[= - \frac{\pi}{3} + (2n + 1)\pi.\]

\[2)\ \cos x = 1\]

\[x = \arccos 1 + 2\pi n = 2\pi n.\]

\[Имеет\ решения\ при:\]

\[\text{tg\ x} > 0\]

\[arctg\ 0 + \pi n < x < \frac{\pi}{2} + \pi n\]

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

\[Ответ:\ \ \frac{\pi}{3} + (2n + 1)\text{π.}\]