ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 1178

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\[1)\arcsin\left( \frac{x}{2} - 3 \right) = \frac{\pi}{6}\]

\[\arcsin\left( \frac{x}{2} - 3 \right) = \arcsin\frac{1}{2}\]

\[\frac{x}{2} - 3 = \frac{1}{2}\]

\[\frac{x}{2} = 3 + \frac{1}{2}\]

\[\frac{x}{2} = \frac{7}{2}\]

\[x = \frac{7}{2} \bullet 2\]

\[x = 7\]

\[Ответ:\ \ x = 7.\]

\[2)\arcsin(3 - 2x) = - \frac{\pi}{4}\]

\[\arcsin(3 - 2x) = - \arcsin\left( \frac{\sqrt{2}}{2} \right)\]

\[\arcsin(3 - 2x) = \arcsin\left( - \frac{\sqrt{2}}{2} \right)\]

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

\[6 - 4x = - \sqrt{2}\]

\[4x = 6 + \sqrt{2}\]

\[x = \frac{6 + \sqrt{2}}{4}\]

\[Ответ:\ \ x = \frac{6 + \sqrt{2}}{4}.\]