Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 762

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\[\mathbf{Н}\mathbf{а\ }отрезке\ \lbrack 0;\ 3\pi\rbrack.\]

\[1)\ 2\cos x + \sqrt{3} = 0\]

\[2\cos x = - \sqrt{3}\]

\[\cos x = - \frac{\sqrt{3}}{2}\]

\[x = \pm \left( \pi - \arccos\frac{\sqrt{3}}{2} \right) + 2\pi n =\]

\[= \pm \left( \pi - \frac{\pi}{6} \right) + 2\pi n =\]

\[= \pm \frac{5\pi}{6} + 2\pi n;\]

\[На\ искомом\ отрезке:\]

\[x_{1} = \frac{5\pi}{6};\]

\[x_{2} = - \frac{5\pi}{6} + 2\pi = \frac{7\pi}{6};\]

\[x_{3} = \frac{5\pi}{6} + 2\pi = \frac{17\pi}{6}.\]

\[2)\ \sqrt{3} - \sin x = \sin x\]

\[2\sin x = \sqrt{3}\]

\[\sin x = \frac{\sqrt{3}}{2}\]

\[x = ( - 1)^{n} \bullet \arcsin\frac{\sqrt{3}}{2} + \pi n =\]

\[= ( - 1)^{n} \bullet \frac{\pi}{3} + \pi n.\]

\[На\ искомом\ отрезке:\]

\[x_{1} = \frac{\pi}{3};\]

\[x_{2} = - \frac{\pi}{3} + \pi = \frac{2\pi}{3};\]

\[x_{3} = \frac{\pi}{3} + 2\pi = \frac{7\pi}{3};\]

\[x_{4} = - \frac{\pi}{3} + 3\pi = \frac{8\pi}{3}.\]

\[3)\ 3\ tg\ x = \sqrt{3}\]

\[tg\ x = \frac{\sqrt{3}}{3}\]

\[x = arctg\frac{\sqrt{3}}{3} + \pi n = \frac{\pi}{6} + \pi n;\]

\[На\ искомом\ отрезке:\]

\[x_{1} = \frac{\pi}{6};\]

\[x_{2} = \frac{\pi}{6} + \pi = \frac{7\pi}{6};\]

\[x_{3} = \frac{\pi}{6} + 2\pi = \frac{13\pi}{6}.\]

\[4)\cos x + 1 = 0\]

\[\cos x = - 1\]

\[x = \pi - \arccos 1 + 2\pi n =\]

\[= \pi + 2\pi n;\]

\[На\ искомом\ отрезке:\]

\[x_{1} = \pi;\]

\[x_{2} = \pi + 2\pi = 3\pi.\]