ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 656

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

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

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

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

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

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

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

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

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

\[6 + 3x =\]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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