Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 1230

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\[1)\cos(4 - 2x) = - \frac{1}{2}\]

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

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

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

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

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

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

\[Ответ:\ \pm \frac{\pi}{3} + 2 - \pi n.\]

\[2)\cos(6 + 3x) = - \frac{\sqrt{2}}{2}\]

\[6 + 3x = \pm \left( \pi - \arccos\frac{\sqrt{2}}{2} \right) +\]

\[+ 2\pi n = \pm \left( \pi - \frac{\pi}{4} \right) +\]

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

\[3x = \pm \frac{3\pi}{4} - 6 + 2\pi n\]

\[x = \frac{1}{3} \bullet \left( \pm \frac{3\pi}{4} - 6 + 2\pi n \right) =\]

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

\[Ответ:\ \pm \frac{\pi}{4} - 2 + \frac{2\pi n}{3}.\]

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

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

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

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

\[+ 2\pi n = \pm \left( \pi - \frac{\pi}{4} \right) +\]

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

\[Первое\ уравнение:\]

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

\[= - \pi + 2\pi n\]

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

\[Второе\ уравнение:\]

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

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

\[Ответ:\ - \frac{\pi}{2} + \pi n;\ \ \frac{\pi}{4} + \pi n.\]

\[4)\ 2\cos\left( \frac{\pi}{3} - 3x \right) - \sqrt{3} = 0\]

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

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

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

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

\[Первое\ уравнение:\]

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

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

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

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

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

\[Второе\ уравнение:\]

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

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

\[x = - \frac{1}{3} \bullet \left( - \frac{\pi}{6} + 2\pi n \right) =\]

\[= \frac{\pi}{18} - \frac{2\pi n}{3}.\]

\[Ответ:\ \ \frac{\pi}{6} - \frac{2\pi n}{3};\ \ \frac{\pi}{18} - \frac{2\pi n}{3}.\]