ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 334

\[1)\ f(x) = 2\sin x + \sin{2x};\left\lbrack 0;\ \frac{3}{2}\pi \right\rbrack:\]

\[f^{'}(x) = 2\cos x + 2\cos{2x}.\]

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

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

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

\[D = 1 + 8 = 9\]

\[\cos x_{1} = \frac{- 1 - 3}{2 \bullet 2} = - 1;\]

\[\cos x_{2} = \frac{- 1 + 3}{2 \bullet 2} = \frac{1}{2};\]

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

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

\[f(0) = 2\sin 0 + \sin 0 = 0 + 0 = 0;\]

\[f\left( \frac{\pi}{3} \right) = 2\sin\frac{\pi}{3} + \sin\frac{2\pi}{3} =\]

\[= \sqrt{3} + \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2};\]

\[f(\pi) = 2\sin\pi + \sin{2\pi} =\]

\[= 0 + 0 = 0;\]

\[f\left( \frac{3\pi}{2} \right) = 2\sin\frac{3\pi}{2} + \sin{3\pi} =\]

\[= - 2 + 0 = - 2.\]

\[Ответ:\ - 2;\ \frac{3\sqrt{3}}{2}.\]

\[2)\ f(x) = 2\cos x + \sin{2x};\lbrack 0;\ \pi\rbrack:\]

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

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

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

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

\[D = 1 + 8 = 9\]

\[\sin x_{1} = \frac{- 1 - 3}{2 \bullet 2} = - 1;\ \]

\[\sin x_{2} = \frac{- 1 + 3}{2 \bullet 2} = \frac{1}{2};\]

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

\[x_{2} = ( - 1)^{n} \bullet \frac{\pi}{6} + \pi n.\]

\[f(0) = 2\cos 0 + \sin 0 = 2 + 0 = 2;\]

\[f\left( \frac{\pi}{6} \right) = 2\cos\frac{\pi}{6} + \sin\frac{\pi}{3} =\]

\[= \sqrt{3} + \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2};\]

\[f\left( \frac{5\pi}{6} \right) = 2\cos\frac{5\pi}{6} + \sin\frac{5\pi}{3} =\]

\[= - \sqrt{3} - \frac{\sqrt{3}}{2} = - \frac{3\sqrt{3}}{2};\]

\[f(\pi) = 2\cos\pi + \sin{2\pi} =\]

\[= - 2 + 0 = - 2.\]

\[Ответ:\ - \frac{3\sqrt{3}}{2};\ \frac{3\sqrt{3}}{2}.\]

\[3)\ f(x) = 3\sin x + 4\cos{2x};\left\lbrack 0;\ \frac{\pi}{2} \right\rbrack:\]

\[f(x) = 3\sin x + 4\left( 1 - 2\sin^{2}x \right);\]

\[f^{'}(x) = 3\cos x - 4 \bullet 2\sin{2x}.\]

\[3\cos x - 8\sin{2x} = 0\]

\[3\cos x - 16\sin x \bullet \cos x = 0\]

\[\cos x \bullet \left( 3 - 16\sin x \right) = 0\]

\[\cos x = 0;\text{\ \ \ }\sin x = \frac{3}{16};\]

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

\[x_{2} = ( - 1)^{n} \bullet \arcsin\frac{3}{16} + \pi n.\]

\[f(0) = 3\sin 0 + 4\cos 0 =\]

\[= 0 + 4 = 4;\]

\[f\left( \arcsin\frac{3}{16} \right) =\]

\[= 3 \bullet \frac{3}{16} + 4\left( 1 - 2 \bullet \left( \frac{3}{16} \right)^{2} \right);\]

\[f\left( \arcsin\frac{3}{16} \right) = \frac{9}{16} + 4 \bullet \frac{119}{128} = \frac{137}{32};\]

\[f\left( \frac{\pi}{2} \right) = 3\sin\frac{\pi}{2} + 4\cos\pi =\]

\[= 3 - 4 = - 1.\]

\[Ответ:\ - 1;\ \frac{137}{32}.\]

\[4)\ f(x) = \sin x + 2\sqrt{2}\cos x;\left\lbrack 0;\ \frac{\pi}{2} \right\rbrack:\]

\[f^{'}(x) = \cos x - 2\sqrt{2}\sin x.\]

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

\[3\left( \frac{1}{3}\cos x - \frac{2\sqrt{2}}{3}\sin x \right) = 0\]

\[\sin\left( x - \arccos\frac{2\sqrt{2}}{3} \right) = 0\]

\[\sin\left( x - \arcsin\frac{1}{3} \right) = 0\]

\[x_{0} = \arccos\frac{2\sqrt{2}}{3} + \pi n;\text{\ \ \ }\]

\[x_{0} = \arcsin\frac{1}{3} + \pi n.\]

\[Значения\ функции:\]

\[f(0) = \sin 0 + 2\sqrt{2}\cos 0 =\]

\[= 0 + 2\sqrt{2} = 2\sqrt{2};\]

\[f\left( x_{0} \right) = \frac{1}{3} + 2\sqrt{2} \bullet \frac{2\sqrt{2}}{3} =\]

\[= \frac{1}{3} + \frac{8}{3} = \frac{9}{3} = 3;\]

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

\[= 1 + 0 = 1.\]

\[Ответ:\ \ 1;\ 3.\]