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МатематикаРешебник по алгебре и начала математического анализа 10 класс Колягин Задание 1230
Задание 1230
\[\boxed{\mathbf{1230}\mathbf{.}}\]
\[1)\cos(4 - 2x) = - \frac{1}{2}\]
\[4 - 2x = \pm \left( \pi - \arccos\frac{1}{2} \right) +\]
\[+ 2\pi n = \pm \left( \pi - \frac{\pi}{3} \right) +\]
\[+ 2\pi n = \pm \frac{2\pi}{3} + 2\pi n\]
\[- 2x = \pm \frac{2\pi}{3} - 4 + 2\pi n\]
\[x = - \frac{1}{2} \bullet \left( \pm \frac{2\pi}{3} - 4 + 2\pi n \right) =\]
\[= \pm \frac{\pi}{3} + 2 - \pi n\]
\[Ответ:\ \pm \frac{\pi}{3} + 2 - \pi n.\]
\[2)\cos(6 + 3x) = - \frac{\sqrt{2}}{2}\]
\[6 + 3x = \pm \left( \pi - \arccos\frac{\sqrt{2}}{2} \right) +\]
\[+ 2\pi n = \pm \left( \pi - \frac{\pi}{4} \right) +\]
\[+ 2\pi n = \pm \frac{3\pi}{4} + 2\pi n\]
\[3x = \pm \frac{3\pi}{4} - 6 + 2\pi n\]
\[x = \frac{1}{3} \bullet \left( \pm \frac{3\pi}{4} - 6 + 2\pi n \right) =\]
\[= \pm \frac{\pi}{4} - 2 + \frac{2\pi n}{3}\]
\[Ответ:\ \pm \frac{\pi}{4} - 2 + \frac{2\pi n}{3}.\]
\[3)\ \sqrt{2}\cos\left( 2x + \frac{\pi}{4} \right) + 1 = 0\]
\[\sqrt{2}\cos\left( 2x + \frac{\pi}{4} \right) = - 1\]
\[\cos\left( 2x + \frac{\pi}{4} \right) = - \frac{1}{\sqrt{2}}\]
\[2x + \frac{\pi}{4} = \pm \left( \pi - \arccos\frac{1}{\sqrt{2}} \right) +\]
\[+ 2\pi n = \pm \left( \pi - \frac{\pi}{4} \right) +\]
\[+ 2\pi n = \pm \frac{3\pi}{4} + 2\pi n\]
\[Первое\ уравнение:\]
\[2x = - \frac{3\pi}{4} - \frac{\pi}{4} + 2\pi n =\]
\[= - \pi + 2\pi n\]
\[x = \frac{1}{2} \bullet ( - \pi + 2\pi n) = - \frac{\pi}{2} + \pi n.\]
\[Второе\ уравнение:\]
\[2x = + \frac{3\pi}{4} - \frac{\pi}{4} + 2\pi n = \frac{\pi}{2} + 2\pi n\]
\[x = \frac{1}{2} \bullet \left( \frac{\pi}{2} + 2\pi n \right) = \frac{\pi}{4} + \pi n.\]
\[Ответ:\ - \frac{\pi}{2} + \pi n;\ \ \frac{\pi}{4} + \pi n.\]
\[4)\ 2\cos\left( \frac{\pi}{3} - 3x \right) - \sqrt{3} = 0\]
\[2\cos\left( \frac{\pi}{3} - 3x \right) = \sqrt{3}\]
\[\cos\left( \frac{\pi}{3} - 3x \right) = \frac{\sqrt{3}}{2}\]
\[\frac{\pi}{3} - 3x = \pm \arccos\frac{\sqrt{3}}{2} + 2\pi n =\]
\[= \pm \frac{\pi}{6} + 2\pi n\]
\[Первое\ уравнение:\]
\[- 3x = - \frac{\pi}{6} - \frac{\pi}{3} + 2\pi n = - \frac{\pi}{6} -\]
\[- \frac{2\pi}{6} + 2\pi n = - \frac{3\pi}{6} + 2\pi n =\]
\[= - \frac{\pi}{2} + 2\pi n\]
\[x = - \frac{1}{3} \bullet \left( - \frac{\pi}{2} + 2\pi n \right) =\]
\[= \frac{\pi}{6} - \frac{2\pi n}{3}.\]
\[Второе\ уравнение:\]
\[- 3x = + \frac{\pi}{6} - \frac{\pi}{3} + 2\pi n = \frac{\pi}{6} -\]
\[- \frac{2\pi}{6} + 2\pi n = - \frac{\pi}{6} + 2\pi n\]
\[x = - \frac{1}{3} \bullet \left( - \frac{\pi}{6} + 2\pi n \right) =\]
\[= \frac{\pi}{18} - \frac{2\pi n}{3}.\]
\[Ответ:\ \ \frac{\pi}{6} - \frac{2\pi n}{3};\ \ \frac{\pi}{18} - \frac{2\pi n}{3}.\]
