Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 623

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\[1)\ 1 + 7\cos^{2}x = 3\sin{2x}\]

\[tg^{2}x + 8 - 6\ tg\ x = 0\]

\[y = tg\ x:\]

\[y^{2} - 6y + 8 = 0\]

\[D = 36 - 32 = 4\]

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

\[y_{2} = \frac{6 + 2}{2} = 4.\]

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

\[x = arctg\ 2 + \pi n.\]

\[2)\ tg\ x = 4\]

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

\[Ответ:\ \ arctg\ 2 + \pi n;\ \ \]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }arctg\ 4 + \pi n.\]

\[2)\ 3 + \sin{2x} = 4\sin^{2}x\]

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

\[y = tg\ x:\]

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

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

\[D = 4 + 12 = 16\]

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

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

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

\[x = - arctg\ 1 + \pi n\]

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

\[2)\ tg\ x = 3\]

\[x = arctg\ 3 + \pi n.\]

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

\[\text{\ \ \ \ \ \ \ \ }arctg\ 3 + \pi n.\]

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

\[y = tg\ x:\]

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

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

\[D = 1 + 8 = 9\]

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

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

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

\[x = - arctg\ 1 + \pi n\]

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

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

\[x = arctg\ 2 + \pi n.\]

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

\[\text{\ \ \ \ \ \ \ \ }arctg\ 2 + \pi n.\]

\[4)\ 3\cos{2x} + \sin^{2}x + 5\sin x \bullet \cos x = 0;\]

\[3 - 2\ tg^{2}x + 5\ tg\ x = 0\]

\[y = tg\ x:\]

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

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

\[D = 25 + 24 = 49\]

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

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

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

\[x = - arctg\frac{1}{2} + \pi n.\]

\[2)\ tg\ x = 3\]

\[x = arctg\ 3 + \pi n.\]

\[Ответ:\ - arctg\frac{1}{2} + \pi n;\ \ \]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }arctg\ 3 + \pi n.\]