Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 753

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

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

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

\[4 - 6x = 1\]

\[6x = 4 - 1\]

\[6x = 3\]

\[x = \frac{1}{2}.\]

\[Ответ:\ \ x = \frac{1}{2}.\]

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

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

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

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

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

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

\[x = \frac{3 \bullet 2 - \sqrt{2}}{4}\]

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

\[Ответ:\ \ \frac{6 - \sqrt{2}}{4}.\]

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

\[\arcsin\frac{x - 2}{4} = - \arcsin\frac{\sqrt{2}}{2}\]

\[\arcsin\frac{x - 2}{4} = \arcsin\left( - \frac{\sqrt{2}}{2} \right)\]

\[\frac{x - 2}{4} = - \frac{\sqrt{2}}{2}\ \ \ \ \ | \bullet 4\]

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

\[x = 2 - 2\sqrt{2}.\]

\[Ответ:\ \ x = 2 - 2\sqrt{2}\]

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

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

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

\[x + 3 = - \sqrt{3}\]

\[x = - 3 - \sqrt{3}.\]

\[Ответ:\ \ x = - 3 - \sqrt{3}.\]