ГДЗ по алгебре и начала математического анализа 11 класс Алимов Задание 657

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\[1)\ 2\sin\left( 3x - \frac{\pi}{4} \right) + 1 = 0\]

\[2\sin\left( \frac{\pi}{2} + \left( 3x - \frac{3\pi}{4} \right) \right) = - 1\]

\[\cos\left( 3x - \frac{3\pi}{4} \right) = - \frac{1}{2}\]

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

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

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

\[= \pm \frac{2\pi}{3} + 2\pi n;\]

\[1)\ 3x = - \frac{2\pi}{3} + \frac{3\pi}{4} + 2\pi n\]

\[3x = - \frac{8\pi}{12} + \frac{9\pi}{12} + 2\pi n\]

\[3x = \frac{\pi}{12} + 2\pi n\]

\[x = \frac{1}{3} \bullet \left( \frac{\pi}{12} + 2\pi n \right) = \frac{\pi}{36} + \frac{2\pi n}{3}.\]

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

\[3x = \frac{8\pi}{12} + \frac{9\pi}{12} + 2\pi n\]

\[3x = \frac{17\pi}{12} + 2\pi n\]

\[x = \frac{1}{3} \bullet \left( \frac{17\pi}{12} + 2\pi n \right)\]

\[x = \frac{17\pi}{36} + \frac{2\pi n}{3}.\]

\[Ответ:\ \ \frac{\pi}{36} + \frac{2\pi n}{3};\ \ \frac{17\pi}{36} + \frac{2\pi n}{3}.\]

\[2)\ 1 - \sin\left( \frac{x}{2} + \frac{\pi}{3} \right) = 0\]

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

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

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

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

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

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

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

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

\[Ответ:\ \ \frac{\pi}{3} + 4\pi n.\]

\[3)\ 3 + 4\sin(2x + 1) = 0\]

\[4\sin(2x + 1) = - 3\]

\[\sin(2x + 1) = - \frac{3}{4}\]

\[2x + 1 = ( - 1)^{n + 1} \bullet \arcsin\frac{3}{4} + \pi n\]

\[2x = ( - 1)^{n + 1} \bullet \arcsin\frac{3}{4} - 1 + \pi n\]

\[x = \frac{1}{2} \bullet \left( - \arcsin\frac{3}{4} - 1 + \pi n \right)\]

\[x = ( - 1)^{n + 1} \bullet \frac{1}{2}\arcsin\frac{3}{4} - \frac{1}{2} + \frac{\text{πn}}{2}.\]

\[Ответ:\ \ \]

\[( - 1)^{n + 1} \bullet \frac{1}{2}\arcsin\frac{3}{4} - \frac{1}{2} + \frac{\text{πn}}{2}.\]

\[4)\ 5\sin(2x - 1) - 2 = 0\]

\[5\sin(2x - 1) = 2\]

\[\sin(2x - 1) = \frac{2}{5}\]

\[2x - 1 = ( - 1)^{n} \bullet \arcsin\frac{2}{5} + \pi n\]

\[2x = ( - 1)^{n} \bullet \arcsin\frac{2}{5} + 1 + \pi n\]

\[x = \frac{1}{2} \bullet \left( ( - 1)^{n} \bullet \arcsin\frac{2}{5} + 1 + \pi n \right)\]

\[x = ( - 1)^{n} \bullet \frac{1}{2}\arcsin\frac{2}{5} + \frac{1}{2} + \frac{\text{πn}}{2}.\]

\[Ответ:\ \ \]

\[( - 1)^{n} \bullet \frac{1}{2}\arcsin\frac{2}{5} + \frac{1}{2} + \frac{\text{πn}}{2}.\]