ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 94

\[1)\ tg^{2}\text{\ x} < 1\]

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

\[(tg\ x + 1)(tg\ x - 1) < 0\]

\[- 1 < tg\ x < 1.\]

\[Ответ:\ \]

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

\[2)\ tg^{2}\text{\ x} \geq 3\]

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

\[\left( tg\ x + \sqrt{3} \right)\left( tg\ x - \sqrt{3} \right) \geq 0\]

\[tg\ x \leq - \sqrt{3}\ \]

\[tg\ x \geq \sqrt{3}.\]

\[Ответ:\ \]

\[- \frac{\pi}{2} + \pi n < x \leq - \frac{\pi}{3} + \pi n;\ \]

\[\frac{\pi}{3} + \pi n \leq x < \frac{\pi}{2} + \pi\text{n.}\]

\[3)\ 3\sin^{2}x + \sin x\cos x > 2\]

\[3\sin^{2}x + \sin x\cos x > 2\sin^{2}x + 2\cos^{2}x\]

\[\sin^{2}x + \sin x\cos x - 2\cos^{2}x > 0\]

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

\[D = 1 + 8 = 9\]

\[\text{tg\ }x_{1} = \frac{- 1 - 3}{2} = - 2;\]

\[\text{tg\ }x_{2} = \frac{- 1 + 3}{2} = 1.\]

\[(tg\ x + 2)(tg\ x - 1) > 0\]

\[\text{tg\ x} < - 2\]

\[\text{tg\ x} > 1.\]

\[- \frac{\pi}{2} + \pi n < x < - arctg\ 2 + \pi n;\text{\ \ \ }\]

\[\frac{\pi}{4} + \pi n < x < \frac{\pi}{2} + \pi n.\]

\[Еще\ решение:\]

\[\cos x = 0;\text{\ \ \ }\sin x = \pm 1;\]

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

\[Ответ:\ \]

\[- \frac{3\pi}{4} + \pi n < x < - arctg\ 2 + \pi\text{n.}\]