Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 689

\[\boxed{\mathbf{689}\mathbf{.}}\]

\[y = \cos\left( x - \frac{\pi}{4} \right):\]

\[2y^{2} - 4ay - 6a^{2} = 0\]

\[D = 16a^{2} + 4 \bullet 2 \bullet 6a^{2} =\]

\[= 16a^{2} + 48a^{2} = 64a^{2}\]

\[y_{1} = \frac{4a - 8a}{2 \bullet 2} = - a;\]

\[y_{2} = \frac{4a + 8a}{2 \bullet 2} = 3a.\]

\[Имеет\ корни\ при:\]

\[\left\{ \begin{matrix} - 1 \leq - a \leq 1 \\ - 1 \leq 3a \leq 1\ \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} - 1 \leq a \leq 1 \\ - \frac{1}{3} \leq a \leq \frac{1}{3} \\ \end{matrix} \right.\ \]

\[При\ - \frac{1}{3} \leq a \leq \frac{1}{3}:\]

\[x_{1} = \frac{\pi}{4} \pm \left( \pi - \arccos a \right) + 2\pi n;\]

\[x_{2} = \frac{\pi}{4} \pm \arccos{3a} + 2\pi n.\]

\[При\ \frac{1}{3} < |a| \leq 1;\]

\[x = \frac{\pi}{4} \pm \left( \pi - \arccos a \right) + 2\pi n.\]