ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 883

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

\[\lg\left( x - \sqrt{2x + 23} \right) > 0.\]

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

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

\[4\sin^{2}x + 4\sin x\cos x = 0\]

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

\[\sin x + \cos x = 0\text{\ \ \ }\]

\[\text{tg\ x} + 1 = 0\text{\ \ \ }\]

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

\[\sin x = 0\]

\[x = \pi n.\]

\[2)\ 4\sin^{3}x \bullet \cos x + 4\sin x \bullet \cos^{3}x =\]

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

\[6\sin^{4}x + 6\sin^{2}x \bullet \cos^{2}x =\]

\[= 6\sin^{2}x \bullet \left( \sin^{2}x + \cos^{2}x \right);\]

\[\cos^{4}x - \sin^{4}x =\]

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

\[3)\ \lg\left( x - \sqrt{2x + 23} \right) > 0\]

\[x - \sqrt{2x + 23} > 1\]

\[\sqrt{2x + 23} < x - 1\]

\[2x + 23 < x^{2} - 2x + 1\]

\[x^{2} - 4x - 22 > 0\]

\[D = 16 + 88 = 104\]

\[x = \frac{4 \pm \sqrt{104}}{2} = \frac{4 \pm 2\sqrt{26}}{2} =\]

\[= 2 \pm \sqrt{26};\]

\[x > 2 + \sqrt{26} \approx 7,1.\]

\[Ответ:\ \ \pi n;\ - \frac{\pi}{4} + \pi n,\ \ \ n \geq 3.\]