Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 857

\[1)\log_{3}\left( \sin{3x} - \sin x \right) =\]

\[= 2\log_{9}\left( 17\sin{2x} \right) - 1\]

\[\log_{3}\left( \sin{3x} - \sin x \right) =\]

\[= \log_{3}\left( 17\sin{2x} \right) - \log_{3}3\]

\[\log_{3}\left( \sin{3x} - \sin x \right) = \log_{3}\frac{17\sin{2x}}{3}\]

\[\sin{3x} - \sin x = \frac{17}{3}\sin{2x}\]

\[2\sin x \bullet \cos{2x} = \frac{17}{3} \bullet 2\sin x \bullet \cos x\]

\[\cos{2x} = \frac{17}{3}\cos x\]

\[2 \bullet \frac{1 + \cos{2x}}{2} - 1 = \frac{17}{3}\cos x\]

\[2\cos^{2}x - \frac{17}{3}\cos x - 1 = 0\]

\[y = \cos x:\]

\[2y^{2} - \frac{17}{3}y - 1 = 0\]

\[6y^{2} - 17y - 3 = 0\]

\[D = 289 + 72 = 361\]

\[y_{1} = \frac{17 - 19}{2 \bullet 6} = - \frac{1}{6};\]

\[y_{2} = \frac{17 + 19}{2 \bullet 6} = 3.\]

\[1)\ \cos x = - \frac{1}{6}\]

\[x = \pi \pm \arccos\frac{1}{6} + 2\pi n.\]

\[2)\ \cos x = 3\]

\[x \in \varnothing.\]

\[Область\ определения:\]

\[\sin{3x} - \sin x > 0;\ \ \sin{2x} > 0\]

\[\sin x \bullet \cos{2x} > 0\text{\ \ \ }\]

\[\sin x \bullet \cos x > 0.\]

\[Ответ:\ \ \pi + \arccos\frac{1}{6} + 2\pi n.\]

\[2)\log_{\sqrt{7}}\left( \sin x - \cos x \right) + 1 =\]

\[= \log_{7}\left( 7 + 3\cos{4x} \right)\]

\[\log_{7}\left( \sin x - \cos x \right)^{2} + \log_{7}7 =\]

\[= \log_{7}\left( 7 + 3\cos{4x} \right)\]

\[\log_{7}{7\left( \sin^{2}x + \cos^{2}x - 2\sin x \bullet \cos x \right)} =\]

\[= \log_{7}\left( 7 + 3\cos{4x} \right)\]

\[7\left( \sin^{2}x + \cos^{2}x - 2\sin x \bullet \cos x \right) =\]

\[= 7 + 3\cos{4x}\]

\[7\left( 1 - \sin{2x} \right) =\]

\[= 7 + 3\left( \cos^{2}{2x} - \sin^{2}{2x} \right)\]

\[7 - 7\sin{2x} =\]

\[= 7 + 3\left( 1 - \sin^{2}{2x} \right) - 3\sin^{2}{2x}\]

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

\[7 - 7y = 7 + 3\left( 1 - y^{2} \right) - 3y^{2}\]

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

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

\[D = 49 + 72 = 121\]

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

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

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

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

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

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

\[x \in \varnothing.\]

\[Область\ определения:\]

\[\sin x - \cos x > 0\]

\[\sin x > \cos x.\]

\[Ответ:\ \ \frac{\pi}{2} + \frac{1}{2}\arcsin\frac{1}{3} + 2\pi n;\ \]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\pi - \frac{1}{2}\arcsin\frac{1}{3} + 2\pi n.\]