ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1620

\[\boxed{\mathbf{1620}\mathbf{.}}\]

\[y = 6\cos^{2}x + 6\sin x - 2;\]

\[y^{'}(x) = 6\left( \cos^{2}x \right)^{'} + 6\left( \sin x \right)^{'} - (2)^{'} =\]

\[= 6 \bullet \left( - \sin x \bullet 2\cos x \right) + 6\cos x =\]

\[= - 12\sin x \bullet \cos x + 6\cos x.\]

\[Стационарные\ точки:\]

\[- 12\sin x \bullet \cos x + 6\cos x = 0\]

\[6\cos x \bullet \left( 1 - 2\sin x \right) = 0.\]

\[1)\ 6\cos x = 0\]

\[\cos x = 0\]

\[x = \arccos 0 + \pi n = \frac{\pi}{2} + \pi n.\]

\[2)\ 2 - 2\sin x = 0\]

\[2\sin x = 1\]

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

\[x = ( - 1)^{n} \bullet \arcsin\frac{1}{2} + \pi n\]

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

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

\[= 6 \bullet 0^{2} + 6 \bullet 1 - 2 = 6 - 2 = 4;\]

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

\[= 6\left( - \sin\pi \right)^{2} + 6\cos\pi - 2 =\]

\[= 6 \bullet 0^{2} - 6 \bullet 1 - 2 = - 8;\]

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

\[= 6 \bullet \frac{3}{4} + 6 \bullet \frac{1}{2} - 2 =\]

\[= 4,5 + 3 - 2 = 5,5;\]

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

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

\[= 6 \bullet \frac{3}{4} + 6 \bullet \frac{1}{2} - 2 = 5,5.\]

\[Ответ:\ \ x = ( - 1)^{n} \bullet \frac{\pi}{6} + \pi n\]