Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 1232

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\[1)\ \left( 1 + \sqrt{2}\cos x \right) \bullet\]

\[\bullet \left( 1 - 4\sin x \bullet \cos x \right) = 0\]

\[Первое\ уравнение:\]

\[1 + \sqrt{2}\cos x = 0\]

\[\sqrt{2}\cos x = - 1\]

\[\cos x = - \frac{1}{\sqrt{2}}\]

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

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

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

\[Второе\ уравнение:\]

\[1 - 4\sin x \bullet \cos x = 0\]

\[2\sin{2x} = 1\]

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

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

\[= ( - 1)^{n} \bullet \frac{\pi}{6} + \pi n\]

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

\[= ( - 1)^{n} \bullet \frac{\pi}{12} + \frac{\text{πn}}{2}.\]

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

\[( - 1)^{n} \bullet \frac{\pi}{12} + \frac{\text{πn}}{2}.\]

\[2)\ \left( 1 - \sqrt{2}\cos x \right) \bullet\]

\[\bullet \left( 1 + 2\sin{2x} \bullet \cos{2x} \right) = 0\]

\[Первое\ уравнение:\]

\[1 - \sqrt{2}\cos x = 0\]

\[\sqrt{2}\cos x = 1\]

\[\cos x = \frac{1}{\sqrt{2}}\]

\[x = \pm \arccos\frac{1}{\sqrt{2}} + 2\pi n =\]

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

\[Второе\ уравнение:\]

\[1 + 2\sin{2x} \bullet \cos{2x} = 0\]

\[\sin{4x} = - 1\]

\[4x = - \arcsin 1 + 2\pi n =\]

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

\[x = \frac{1}{4} \bullet \left( - \frac{\pi}{2} + 2\pi n \right) = - \frac{\pi}{8} + \frac{\text{πn}}{2}.\]

\[Ответ:\ \pm \frac{\pi}{4} + 2\pi n;\ \ - \frac{\pi}{8} + \frac{\text{πn}}{2}.\]