Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 1055

\[y = \frac{2}{3}\cos\left( 3x - \frac{\pi}{6} \right);\ \ x_{0} = \frac{\pi}{3}:\]

\[y^{'}(x) = \frac{2}{3} \bullet 3 \bullet \left( - \sin\left( 3x - \frac{\pi}{6} \right) \right) =\]

\[= - 2\sin\left( 3x - \frac{\pi}{6} \right);\]

\[y^{'}\left( \frac{\pi}{3} \right) = - 2\sin\left( \frac{3\pi}{3} - \frac{\pi}{6} \right) =\]

\[= - 2\sin\left( \pi - \frac{\pi}{6} \right) = - 2\sin\frac{\pi}{6} =\]

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

\[a = arctg\ k = arctg( - 1) = - \frac{\pi}{4}.\]

\[Ответ:\ \ a = - \frac{\pi}{4}\text{.\ }\]