Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1417

\[\boxed{\mathbf{1417}\mathbf{.}}\]

\[\mathbf{Н}\mathbf{а\ отрез}ке\ \lbrack - 3\pi;\ \pi\rbrack.\]

\[1)\ 2\cos x - \sqrt{3} < 0\]

\[2\cos x < \sqrt{3}\]

\[\cos x < \frac{\sqrt{3}}{2}\]

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

\[= \pm \frac{\pi}{6} + 2\pi n;\]

\[y = \cos x\ и\ y = \frac{\sqrt{3}}{2}:\]

\[- 3\pi \leq x_{1} < - \frac{13\pi}{6};\]

\[- \frac{11\pi}{6} < x_{2} < - \frac{\pi}{6};\]

\[\frac{\pi}{6} < x_{3} \leq \pi.\]

\[2)\ \sqrt{2}\sin x + 1 \geq 0\]

\[\sqrt{2}\sin x \geq - 1\]

\[\sin x \geq - \frac{\sqrt{2}}{2}\]

\[x = ( - 1)^{n + 1} \bullet \arcsin\frac{\sqrt{2}}{2} + \pi n =\]

\[= ( - 1)^{n + 1} \bullet \frac{\pi}{4} + \pi n;\]

\[y = \sin x\ и\ y = - \frac{\sqrt{2}}{2}:\]

\[- 3\pi \leq x_{1} \leq - \frac{11\pi}{4};\]

\[- \frac{9\pi}{4} \leq x_{2} \leq - \frac{3\pi}{4};\]

\[- \frac{\pi}{4} \leq x_{3} \leq \pi.\]

\[3)\ \sqrt{3} + tg\ x \leq 0\]

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

\[x = - arctg\ \sqrt{3} + \pi n = - \frac{\pi}{3} + \pi n;\]

\[y = tg\ x\ и\ y = - \sqrt{3}:\]

\[- \frac{5\pi}{2} < x_{1} \leq - \frac{7\pi}{3};\]

\[- \frac{3\pi}{2} < x_{2} \leq - \frac{4\pi}{3};\]

\[- \frac{\pi}{2} < x_{3} \leq - \frac{\pi}{3};\]

\[\frac{\pi}{2} < x_{4} \leq \frac{2\pi}{3}.\]

\[4)\ 3\ tg\ x - 2 > 0\]

\[3\ tg\ x > 2\]

\[tg\ x > \frac{2}{3}\]

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

\[y = tg\ x\ и\ y = \frac{2}{3}:\]

\[\text{arctg}\frac{2}{3} - 3\pi < x_{1} < - \frac{5\pi}{2};\]

\[\text{arctg}\frac{2}{3} - 2\pi < x_{2} < - \frac{3\pi}{2};\]

\[\text{arctg}\frac{2}{3} - \pi < x_{3} < - \frac{\pi}{2};\]

\[\text{arctg}\frac{2}{3} < x_{4} < \frac{\pi}{2}.\]