Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 1065

\[y = 2 - 3\sin x + 4\cos x;\]

\[на\ \left\lbrack - \frac{4\pi}{3};\ \frac{2\pi}{3} \right\rbrack:\]

\[1)\ y^{'}(x) =\]

\[= 0 - 3\cos x + 4\left( - \sin x \right) = 0;\]

\[4\sin x = - 3\cos x\ \ \ \ \ |\ :\cos x\]

\[4\ tg\ x = - 3\]

\[tg\ x = - \frac{3}{4}.\]

\[2)\ \cos(x) = \sqrt{\frac{1}{tg^{2}\ x + 1}} =\]

\[= \sqrt{\frac{1}{\frac{9}{16} + \frac{16}{16}}} = \sqrt{1\ :\frac{25}{16}} =\]

\[= \sqrt{\frac{16}{25}} = \pm \frac{4}{5};\]

\[\sin(x) = \sqrt{1 - \cos^{2}(x)} =\]

\[= \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{25}{25} - \frac{16}{25}} =\]

\[= \sqrt{\frac{9}{25}} = \mp \frac{3}{5}.\]

\[3)\ Полный\ период\ функции:\]

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

\[= \frac{6\pi}{3} = 2\pi.\]

\[4)\ y\left( x_{1} \right) = 2 - 3 \bullet \left( - \frac{3}{5} \right) + 4 \bullet \frac{4}{5} =\]

\[= 2 + \frac{9 + 16}{5} = 2 + \frac{25}{5} = 7;\]

\[y\left( x_{2} \right) = 2 - 3 \bullet \frac{3}{5} + 4 \bullet \left( - \frac{4}{5} \right) =\]

\[= 2 - \frac{9 + 16}{5} = 2 - \frac{25}{5} = - 3.\]

\[Ответ:\ - 3;\ 7.\]