Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 881

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

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

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

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

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

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

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

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

\[\frac{1}{y} - \frac{1}{3} = 2 + 2y\ \ \ \ \ | \bullet 3y\]

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

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

\[D = 49 + 72 = 121\]

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

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

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

\[x \in \varnothing.\]

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

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

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

\[Ответ:\ \ \frac{1}{2}\left( ( - 1)^{n}\arcsin\frac{1}{3} + \pi n \right).\]

\[2)\ \frac{\sin{3x} - 2\cos{2x} - 1}{1 - \sin x} = \left| \sin x \right|\]

\[\sin x \geq 0:\]

\[\frac{\left( 4\sin^{2}x - 3 \right)\left( 1 - \sin x \right)}{1 - \sin x} - \sin x = 0\]

\[4\sin^{2}x - \sin x - 3 = 0\]

\[D = 1 + 48 = 49\]

\[\sin x_{1} = \frac{1 - 7}{2 \bullet 4} = - \frac{3}{4};\]

\[\sin x_{2} = \frac{1 + 7}{2 \bullet 4} = 1.\]

\[\sin x = 1\text{\ \ \ }\]

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

\[\sin x \leq 0:\]

\[4\sin^{2}x + \sin x - 3 = 0\]

\[D = 1 + 48 = 49\]

\[\sin x_{1} = \frac{- 1 - 7}{2 \bullet 4} = - 1;\]

\[\sin x_{2} = \frac{- 1 + 7}{2 \bullet 4} = \frac{3}{4}.\]

\[\sin x = - 1\]

\[x = \frac{3\pi}{2} + 2\pi n.\]

\[Область\ определения:\]

\[1 - \sin x \neq 0\]

\[\sin x \neq 1.\]

\[Ответ:\ \ \frac{3\pi}{2} + 2\pi n.\]