ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1601

\[\boxed{\mathbf{1601}\mathbf{.}}\]

\[\frac{2\sin x}{\cos x - \cos{3x}} - \frac{1}{3} = 4\sin^{2}\left( x + \frac{\pi}{4} \right)\]

\[\frac{2\sin x}{- 2 \bullet \sin{\frac{x + 3x}{2} \bullet \sin\frac{x - 3x}{2}}} - \frac{1}{3} =\]

\[= 4 \bullet \frac{1 - \cos\left( 2x + \frac{\pi}{2} \right)}{2}\]

\[\frac{\sin x}{\sin{2x} \bullet \sin x} - \frac{1}{3} = 2\left( 1 + \sin{2x} \right)\]

\[\frac{1}{\sin{2x}} - \frac{1}{3} - \left( 2 + 2\sin{2x} \right) = 0\]

\[\frac{3 - \sin{2x} - 3\sin{2x} \bullet \left( 2 + 2\sin{2x} \right)}{3\sin{2x}} = 0\]

\[\frac{3 - \sin{2x} - 6\sin{2x} - 6\sin^{2}{2x}}{3\sin{2x}} = 0\]

\[6\sin^{2}{2x} + 7\sin{2x} - 3 = 0\]

\[y = \sin{2x}:\]

\[6y^{2} + 7y - 3 = 0\]

\[D = 49 + 72 = 121\]

\[y_{1} = \frac{- 7 - 11}{2 \bullet 6} = - \frac{18}{12} = - \frac{3}{2};\]

\[y_{2} = \frac{- 7 + 11}{2 \bullet 6} = \frac{4}{12} = \frac{1}{3}.\]

\[1)\ \sin{2x} = - \frac{3}{2}\]

\[корней\ нет.\]

\[2)\ \sin{2x} = \frac{1}{3}\]

\[2x = ( - 1)^{n} \bullet \arcsin\frac{1}{3} + \pi n\]

\[x = ( - 1)^{n} \bullet \frac{1}{2}\arcsin\frac{1}{3} + \frac{\text{πn}}{2}.\]

\[Имеет\ смысл\ при:\]

\[\sin x \neq 0\]

\[x \neq \arcsin 0 + \pi n \neq \pi n.\]

\[\sin{3x} \neq 0\]

\[3x \neq \arcsin 0 + \pi n \neq \pi n\]

\[x \neq \frac{\text{πn}}{3}.\]

\[Ответ:\ \ ( - 1)^{n} \bullet \frac{1}{2}\arcsin\frac{1}{3} + \frac{\text{πn}}{2}.\]