Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 117

\[\lbrack 0;\ 3\pi\rbrack.\]

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

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

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

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

\[Ответ:\ \ \frac{5\pi}{6};\ \frac{7\pi}{6};\ \frac{17\pi}{6}.\]

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

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

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

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

\[Ответ:\ \ \frac{\pi}{3};\ \frac{2\pi}{3};\ \frac{7\pi}{3};\ \frac{8\pi}{3}.\]

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

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

\[x = \frac{\pi}{6} + \pi\text{n.}\]

\[Ответ:\ \ \frac{\pi}{6};\ \frac{7\pi}{6};\ \frac{13\pi}{6}.\]

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

\[\cos x = - 1\]

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

\[Ответ:\ \ \pi;\ 3\pi.\]