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МатематикаРешебник по алгебре и начала математического анализа 10 класс Колягин Задание 1231
Задание 1231
\[\boxed{\mathbf{1231}\mathbf{.}}\]
\[1)\ 2\sin\left( 3x - \frac{\pi}{4} \right) + 1 = 0\]
\[2\sin\left( \frac{\pi}{2} + \left( 3x - \frac{3\pi}{4} \right) \right) = - 1\]
\[\cos\left( 3x - \frac{3\pi}{4} \right) = - \frac{1}{2}\]
\[3x - \frac{3\pi}{4} = \pm \left( \pi - \arccos\frac{1}{2} \right) +\]
\[+ 2\pi n = \pm \left( \pi - \frac{\pi}{3} \right) + 2\pi n =\]
\[= \pm \frac{2\pi}{3} + 2\pi n\]
\[Первое\ уравнение:\]
\[3x = - \frac{2\pi}{3} + \frac{3\pi}{4} + 2\pi n =\]
\[= - \frac{8\pi}{12} + \frac{9\pi}{12} + 2\pi n = \frac{\pi}{12} + 2\pi n\]
\[x = \frac{1}{3} \bullet \left( \frac{\pi}{12} + 2\pi n \right) = \frac{\pi}{36} + \frac{2\pi n}{3}.\]
\[Второе\ уравнение:\]
\[3x = + \frac{2\pi}{3} + \frac{3\pi}{4} + 2\pi n = \frac{8\pi}{12} +\]
\[+ \frac{9\pi}{12} + 2\pi n = \frac{17\pi}{12} + 2\pi n\]
\[x = \frac{1}{3} \bullet \left( \frac{17\pi}{12} + 2\pi n \right) = \frac{17\pi}{36} +\]
\[+ \frac{2\pi n}{3}.\]
\[Ответ:\ \ \frac{\pi}{36} + \frac{2\pi n}{3};\ \ \frac{17\pi}{36} + \frac{2\pi n}{3}.\]
\[2)\ 1 - \sin\left( \frac{x}{2} + \frac{\pi}{3} \right) = 0\]
\[\sin\left( \frac{x}{2} + \frac{\pi}{3} \right) = 1\]
\[\frac{x}{2} + \frac{\pi}{3} = \arcsin 1 + 2\pi n =\]
\[= \frac{\pi}{2} + 2\pi n\]
\[\frac{x}{2} = \frac{\pi}{2} - \frac{\pi}{3} + 2\pi n = \frac{3\pi}{6} - \frac{2\pi}{6} +\]
\[+ 2\pi n = \frac{\pi}{6} + 2\pi n\]
\[x = 2 \bullet \left( \frac{\pi}{6} + 2\pi n \right) = \frac{\pi}{3} + 4\pi n.\]
\[Ответ:\ \ \frac{\pi}{3} + 4\pi n.\]
\[3)\ 3 + 4\sin(2x + 1) = 0\]
\[4\sin(2x + 1) = - 3\]
\[\sin(2x + 1) = - \frac{3}{4}\]
\[2x + 1 = ( - 1)^{n + 1} \bullet \arcsin\frac{3}{4} +\]
\[+ \text{πn}\]
\[2x = ( - 1)^{n + 1} \bullet \arcsin\frac{3}{4} - 1 +\]
\[+ \text{πn}\]
\[x = \frac{1}{2} \bullet \left( - \arcsin\frac{3}{4} - 1 + \pi n \right) =\]
\[= ( - 1)^{n + 1} \bullet \frac{1}{2}\arcsin\frac{3}{4} - \frac{1}{2} + \frac{\text{πn}}{2}.\]
\[Ответ:\ \ ( - 1)^{n + 1} \bullet \frac{1}{2}\arcsin\frac{3}{4} -\]
\[- \frac{1}{2} + \frac{\text{πn}}{2}.\]
\[4)\ 5\sin(2x - 1) - 2 = 0\]
\[5\sin(2x - 1) = 2\]
\[\sin(2x - 1) = \frac{2}{5}\]
\[2x - 1 = ( - 1)^{n} \bullet \arcsin\frac{2}{5} + \pi n\]
\[2x = ( - 1)^{n} \bullet \arcsin\frac{2}{5} + 1 + \pi n\]
\[x = \frac{1}{2} \bullet\]
\[\bullet \left( ( - 1)^{n} \bullet \arcsin\frac{2}{5} + 1 + \pi n \right) =\]
\[= ( - 1)^{n} \bullet \frac{1}{2}\arcsin\frac{2}{5} + \frac{1}{2} + \frac{\text{πn}}{2}.\]
\[Ответ:\ \ ( - 1)^{n} \bullet \frac{1}{2}\arcsin\frac{2}{5} +\]
\[+ \frac{1}{2} + \frac{\text{πn}}{2}.\]
