Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1378

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\[tg^{2}\ x - 8\ tg\ x + 7 = 0\]

\[y = tg\ x:\]

\[y^{2} - 8y + 7 = 0\]

\[D = 64 - 28 = 36\]

\[y_{1} = \frac{8 - 6}{2} = 1;\]

\[y_{2} = \frac{8 + 6}{2} = 7.\]

\[1)\ tg\ x = 1\]

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

\[2)\ tg\ x = 7\]

\[x = arctg\ 7 + \pi n.\]

\[Ответ:\ \ \frac{\pi}{4} + \pi n;\ \ arctg\ 7 + \pi n.\]

\[2)\ 3\sin^{2}x - 2\sin x \bullet \cos x = 1\]

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

\[D = 4 + 8 = 12 = 4 \bullet 3\]

\[tg\ x = \frac{2 \pm \sqrt{12}}{2 \bullet 2} =\]

\[= \frac{2 \pm 2\sqrt{3}}{4} = \frac{1 \pm \sqrt{3}}{2}\]

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

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