Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 1194

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\[1)\ tg^{2}\ x = 2\]

\[tg\ x = \pm \sqrt{2}\]

\[x = \pm arctg\ \sqrt{2} + \pi n\]

\[Ответ:\ \pm arctg\ \sqrt{2} + \pi n.\]

\[2)\ tg\ x = ctg\ x\]

\[tg\ x = \frac{1}{\text{tg\ x}}\]

\[tg^{2}\ x = 1\]

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

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

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

\[2\cos{2x} = 0\]

\[\cos{2x} = 0\]

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

\[x = \frac{1}{2} \bullet \left( \frac{\pi}{2} + \pi n \right) = \frac{\pi}{4} + \frac{\text{πn}}{2}\]

\[Ответ:\ \ \frac{\pi}{4} + \frac{\text{πn}}{2}.\]

\[3)\ tg^{2}\ x - 3\ tg\ x - 4 = 0\]

\[Пусть\ y = tg\ x:\]

\[y^{2} - 3y - 4 = 0\]

\[D = 3^{2} + 4 \bullet 4 = 9 + 16 = 25\]

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

\[\text{\ \ }y_{2} = \frac{3 + 5}{2} = 4.\]

\[Первое\ уравнение:\]

\[tg\ x = - 1\]

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

\[Второе\ уравнение:\]

\[tg\ x = 4\]

\[x = arctg\ 4 + \pi n.\]

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

\[4)\ tg^{2}\ x - tg\ x + 1 = 0\]

\[Пусть\ y = tg\ x:\]

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

\[D = 1^{2} - 4 = 1 - 4 = - 3\]

\[D < 0 - \ корней\ нет.\]

\[Ответ:\ \ корней\ нет.\]