ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 1231

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

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

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

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

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

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

\[Первое\ уравнение:\]

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

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

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

\[Второе\ уравнение:\]

\[3x = + \frac{2\pi}{3} + \frac{3\pi}{4} + 2\pi n = \frac{8\pi}{12} +\]

\[+ \frac{9\pi}{12} + 2\pi n = \frac{17\pi}{12} + 2\pi n\]

\[x = \frac{1}{3} \bullet \left( \frac{17\pi}{12} + 2\pi n \right) = \frac{17\pi}{36} +\]

\[+ \frac{2\pi n}{3}.\]

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

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

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

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

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

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

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

\[x = 2 \bullet \left( \frac{\pi}{6} + 2\pi n \right) = \frac{\pi}{3} + 4\pi n.\]

\[Ответ:\ \ \frac{\pi}{3} + 4\pi n.\]

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

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

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

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

\[+ \text{πn}\]

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

\[+ \text{πn}\]

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

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

\[Ответ:\ \ ( - 1)^{n + 1} \bullet \frac{1}{2}\arcsin\frac{3}{4} -\]

\[- \frac{1}{2} + \frac{\text{πn}}{2}.\]

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

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

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

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

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

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

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

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

\[Ответ:\ \ ( - 1)^{n} \bullet \frac{1}{2}\arcsin\frac{2}{5} +\]

\[+ \frac{1}{2} + \frac{\text{πn}}{2}.\]