ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 852

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\[\cos x + \sin x - \sqrt{2}\sin{5x} = 0\]

\[\sin\left( \frac{\pi}{2} - x \right) + \sin x - \sqrt{2}\sin{5x} = 0\]

\[2 \bullet \sin\frac{\frac{\pi}{2} - x + x}{2} \bullet \cos\frac{\frac{\pi}{2} - x - x}{2} - \sqrt{2}\sin{5x} = 0\]

\[2\sin\frac{\pi}{4} \bullet \cos\left( \frac{\pi}{4} - x \right) - \sqrt{2}\sin{5x} = 0\]

\[\sqrt{2} \bullet \left( \cos\left( \frac{\pi}{4} - x \right) - \sin{5x} \right) = 0\]

\[\sqrt{2} \bullet \left( \sin\left( \frac{\pi}{2} - \frac{\pi}{4} + x \right) - \sin{5x} \right) = 0\]

\[\sin\left( \frac{\pi}{4} + x \right) - \sin{5x} = 0\]

\[2 \bullet \sin\frac{\frac{\pi}{4} + x - 5x}{2} \bullet \cos\frac{\frac{\pi}{4} + x + 5x}{2} = 0\]

\[\sin\left( \frac{\pi}{8} - 2x \right) \bullet \cos\left( \frac{\pi}{8} + 3x \right) = 0\]

\[- \sin\left( 2x - \frac{\pi}{8} \right) \bullet \cos\left( 3x + \frac{\pi}{8} \right) = 0\]

\[1)\ \sin\left( 2x - \frac{\pi}{8} \right) = 0\]

\[2x - \frac{\pi}{8} = \arcsin 0 + \pi n = \pi n\]

\[2x = \frac{\pi}{8} + \pi n\]

\[x = \frac{1}{2} \bullet \left( \frac{\pi}{8} + \pi n \right) = \frac{\pi}{16} + \frac{\text{πn}}{2}.\]

\[2)\ \cos\left( 3x + \frac{\pi}{8} \right) = 0\]

\[3x + \frac{\pi}{8} = \arccos 0 + \pi n\]

\[3x + \frac{\pi}{8} = \frac{\pi}{2} + \pi n\]

\[3x = \frac{\pi}{2} - \frac{\pi}{8} + \pi n = \frac{3\pi}{8} + \pi n\]

\[x = \frac{1}{3} \bullet \left( \frac{3\pi}{8} + \pi n \right) = \frac{\pi}{8} + \frac{\text{πn}}{3}.\]

\[Ответ:\ \ \frac{\pi}{16} + \frac{\text{πn}}{2}\text{\ \ }\frac{\pi}{8} + \frac{\text{πn}}{3}.\]

\[y = \cos\left( x - \frac{\pi}{4} \right):\]

\[- \sqrt{2} \bullet y - 10y^{2} + 6 = 0\]

\[10y^{2} + \sqrt{2}y - 6 = 0\]

\[D = 2 + 240 = 242 = 121 \bullet 2\]

\[y_{1} = \frac{- \sqrt{2} - 11\sqrt{2}}{2 \bullet 10} =\]

\[= - \frac{12\sqrt{2}}{20} = - \frac{3\sqrt{2}}{5}\]

\[y_{2} = \frac{- \sqrt{2} + 11\sqrt{2}}{2 \bullet 10} = \frac{10\sqrt{2}}{20} = \frac{\sqrt{2}}{2}.\]

\[1)\ \cos\left( x - \frac{\pi}{4} \right) = - \frac{3\sqrt{2}}{5}\]

\[x - \frac{\pi}{4} = \pm \left( \pi - \arccos\frac{3\sqrt{2}}{5} \right) + 2\pi n\]

\[x = \frac{\pi}{4} \pm \left( \pi - \arccos\frac{3\sqrt{2}}{5} \right) + 2\pi n.\]

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

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

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

\[x_{1} = - \frac{\pi}{4} + \frac{\pi}{4} + 2\pi n = 2\pi n\]

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

\[Ответ:\ \ \]

\[\frac{\pi}{4} \pm \left( \pi - \arccos\frac{3\sqrt{2}}{5} \right) + 2\pi n\ \ \]

\[2\pi n\ \ \frac{\pi}{2} + 2\pi n.\ \]