Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 1274

\[\boxed{\mathbf{1274}\mathbf{.}}\]

\[\sin{2x} - 2a\sqrt{2} \bullet \left( \sin x + \cos x \right) +\]

\[+ 1 - 6a^{2} = 0\]

\[\sin\left( \frac{\pi}{2} + 2x - \frac{\pi}{2} \right) - 2a\sqrt{2} \bullet\]

\[\bullet \left( \sin\left( \frac{\pi}{2} + x - \frac{\pi}{2} \right) + \cos x \right) +\]

\[+ 1 - 6a^{2} = 0\]

\[1 + \cos\left( 2x - \frac{\pi}{2} \right) - 2a\sqrt{2} \bullet\]

\[\bullet \left( \cos\left( x - \frac{\pi}{2} \right) + \cos x \right) -\]

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

\[2 \bullet \frac{1 + \cos\left( 2x - \frac{\pi}{2} \right)}{2} - 2a\sqrt{2} \bullet 2 \bullet\]

\[\bullet \cos\frac{x - \frac{\pi}{2} + x}{2} \bullet \cos\frac{x - \frac{\pi}{2} - x}{2} -\]

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

\[2\cos^{2}\left( x - \frac{\pi}{4} \right) - 4a\sqrt{2} \bullet\]

\[\bullet \cos\left( x - \frac{\pi}{4} \right) \bullet \cos\frac{\pi}{4} - 6a^{2} = 0\]

\[2\cos^{2}\left( x - \frac{\pi}{4} \right) - 4a \bullet\]

\[\bullet \cos\left( x - \frac{\pi}{4} \right) - 6a^{2} = 0\]

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

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

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

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

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

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

\[Уравнение\ имеет\ корни\ при:\]

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

\[Значения\ корней\ уравнения\ \]

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

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

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

\[Значения\ уравнения\ при\]

\[\ \frac{1}{3} < |a| \leq 1\]

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