ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1603

\[\boxed{\mathbf{1603}\mathbf{.}}\]

\[\sin^{4}x + \sin^{4}\left( x + \frac{\pi}{4} \right) = \sin^{2}\frac{25\pi}{6}\]

\[\sin^{4}x + \left( \sin^{2}\left( x + \frac{\pi}{4} \right) \right)^{2} =\]

\[= \sin^{2}\left( 4\pi + \frac{\pi}{6} \right)\]

\[\sin^{4}x + \left( \frac{1 - \cos\left( 2x + \frac{\pi}{2} \right)}{2} \right)^{2} = \sin^{2}\frac{\pi}{6}\]

\[\sin^{4}x + \frac{\left( 1 + \sin{2x} \right)^{2}}{4} = \left( \frac{1}{2} \right)^{2}\]

\[4\left( \sin^{2}x \right)^{2} + 1 + 2\sin{2x} + \sin^{2}{2x} = 1\]

\[4 \bullet \left( \frac{1 - \cos{2x}}{2} \right)^{2} + 2\sin{2x} + \sin^{2}{2x} = 0\]

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

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

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

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

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

\[\sin\frac{\pi}{4} \bullet \sin{2x} - \cos\frac{\pi}{4} \bullet \cos{2x} = - \frac{\sqrt{2}}{2}\]

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

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

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

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

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

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

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

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

\[2x_{2} = 2\pi n\]

\[x_{2} = \pi n.\]

\[Имеет\ решения\ при:\]

\[\lg\left( x - \sqrt{2x + 24} \right) > 0\]

\[\lg\left( x - \sqrt{2x + 24} \right) > \lg 10^{0}\]

\[x - \sqrt{2x + 24} > 1\]

\[x - 1 > \sqrt{2x + 24}\]

\[x^{2} - 2x + 1 > 2x + 24\]

\[x^{2} - 4x - 23 > 0\]

\[D = 16 + 92 = 108 = 4 \bullet 27\]

\[x = \frac{4 \pm \sqrt{108}}{2} = \frac{4 \pm 2\sqrt{27}}{2} =\]

\[= 2 \pm \sqrt{27}\]

\[(x - \left( 2 - \sqrt{27} \right))(x - \left( 2 + \sqrt{27} \right) > 0\]

\[x < 2 - \sqrt{27}\text{\ \ }и\ \ x > 2 + \sqrt{27}.\]

\[x - \sqrt{2x + 24} > 0\]

\[x > \sqrt{2x + 24}\]

\[x^{2} > 2x + 24\]

\[x^{2} - 2x - 24 > 0\]

\[D = 4 + 96 = 100\]

\[x_{1} = \frac{2 - 10}{2} = - 4;\]

\[x_{2} = \frac{2 + 10}{2} = 6;\]

\[(x + 4)(x - 6) > 0\]

\[x < - 4\ \ и\ \ x > 6.\]

\[Имеет\ смысл\ при:\]

\[2x + 24 > 0\]

\[x + 12 > 0\]

\[x > - 12.\]

\[x - 1 > 0\]

\[x > 1.\]

\[Имеем:\ \ \]

\[x > 6.\]

\[Ответ:\ \ x = \pi n;\ \ \]

\[x = - \frac{\pi}{4} + \pi n;\ \ n > 2.\ \ \]