ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 217

\[1)\ f(x) =\]

\[= 5\left( \sin x - \cos x \right) + \sqrt{2}\cos{5x};\]

\[f^{'}(x) =\]

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

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

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

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

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

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

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

\[x = \frac{\pi}{16} + \frac{\pi n}{2}.\]

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

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

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

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

\[Ответ:\ \ \frac{\pi}{16} + \frac{\pi n}{2};\ \frac{\pi}{8} + \frac{\pi n}{3}.\]

\[2)\ f(x) =\]

\[= 1 - \cos{2x} + \sin x - \cos x - x;\]

\[f^{'}(x) =\]

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

\[\cos x + \sin x = 1 - 4\sin x\cos x\]

\[y = \cos x + \sin x:\]

\[y^{2} = 1 + 2\sin x\cos x\]

\[2\sin x\cos x = y^{2} - 1\]

\[y = 1 - 2\left( y^{2} - 1 \right)\]

\[2y^{2} + y - 3 = 0\]

\[D = 1 + 24 = 25\]

\[y_{1} = \frac{- 1 - 5}{2 \bullet 2} = - \frac{3}{2};\text{\ \ }\]

\[y_{2} = \frac{- 1 + 5}{2 \bullet 2} = 1.\]

\[1)\ \sin x + \cos x = - \frac{3}{2}\]

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

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

\[x \in \varnothing.\]

\[2)\ \sin x + \cos x = - \frac{3}{2}\]

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

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

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

\[x = 2\pi n.\]

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

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

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