Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 591

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\[1)\sin{3x} = 1\]

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

\[x = \frac{1}{3} \bullet \left( \frac{\pi}{2} + 2\pi n \right)\]

\[x = \frac{\pi}{6} + \frac{2\pi n}{3}\]

\[Ответ:\ \ x = \frac{\pi}{6} + \frac{2\pi n}{3}.\]

\[2)\sin{2x} = - 1\]

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

\[= - \frac{\pi}{2} + 2\pi n\]

\[x = \frac{1}{2} \bullet \left( - \frac{\pi}{2} + 2\pi n \right)\]

\[x = - \frac{\pi}{4} + \pi n\]

\[Ответ:\ x = - \frac{\pi}{4} + \pi n.\]

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

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

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

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

\[x = 3 \bullet \left( ( - 1)^{n + 1} \bullet \frac{\pi}{4} + \pi n \right)\]

\[x = ( - 1)^{n + 1} \bullet \frac{3\pi}{4} + 3\pi n\]

\[Ответ:\ \ x = ( - 1)^{n + 1} \bullet \frac{3\pi}{4} + 3\pi n.\ \]

\[4)\ 2\sin\frac{x}{2} = \sqrt{3}\]

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

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

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

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

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

\[Ответ:\ \ x = ( - 1)^{n} \bullet \frac{2\pi}{3} + 2\pi n.\]

\[5)\sin\left( x + \frac{3\pi}{4} \right) = 0\]

\[x + \frac{3\pi}{4} = \arcsin 0 + \pi n = \pi n\]

\[x = \pi n = - \frac{3\pi}{4}\]

\[Ответ:\ x = - \frac{3\pi}{4} + \pi n\]

\[6)\sin\left( 2x + \frac{\pi}{2} \right) = 0\]

\[2x + \frac{\pi}{2} = \arcsin 0 + \pi n = \pi n\]

\[2x = \pi n - \frac{\pi}{2}\]

\[x = \frac{1}{2} \bullet \left( \pi n - \frac{\pi}{2} \right)\]

\[x = \frac{\text{πn}}{2} - \frac{\pi}{4}\]

\[Ответ:\ x = - \frac{\pi}{4} + \frac{\text{πn}}{2}.\]