ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 1062

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

\[y^{'}(x) = 2\cos x + 2\cos{2x} = 0\]

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

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

\[D = 1 + 8 = 9\]

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

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

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

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

\[y(0) = 2\sin 0 + \sin 0 =\]

\[= 2 \bullet 0 + 0 = 0;\]

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

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

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

\[= 2 \bullet 0 + 0 = 0;\]

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

\[= 2 \bullet ( - 1) + 0 = - 2.\]

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

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

\[y^{'}(x) = 2\cos x - 2\sin{2x} = 0;\]

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

\[\cos x - 2\sin x \bullet \cos x = 0\]

\[\cos x \bullet \left( 1 - 2\sin x \right) = 0\]

\[\cos x_{1} = 0\]

\[x_{1} = \pi n.\ \ \ \]

\[\sin x_{2} = \frac{1}{2}\]

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

\[y(0) = 2\sin 0 + \cos 0 =\]

\[= 2 \bullet 0 + 1 = 1;\]

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

\[= 2 \bullet \frac{1}{2} + \frac{1}{2} = 1,5;\]

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

\[= 2 \bullet 1 - 1 = 1.\]

\[Ответ:\ \ 1;\ 1,5.\]