ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 856

\[1)\log_{\sqrt{6}}\text{ctg\ x} = 1 + \log_{6}\left( \frac{3}{2} - \cos{2x} \right)\]

\[\log_{6}{\text{ct}g^{2}\text{\ x}} = \log_{6}6 + \log_{6}\left( 1,5 - \cos{2x} \right)\]

\[\log_{6}{\text{ct}g^{2}\text{\ x}} = \log_{6}\left( 9 - 6\cos{2x} \right)\]

\[\text{ct}g^{2}\ x = 9 - 6\cos{2x}\]

\[\frac{\cos^{2}x}{\sin^{2}x} = 9 - 6\left( \cos^{2}x - \sin^{2}x \right)\]

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

\[= 9 - 6\cos^{2}x + 6\left( 1 - \cos^{2}x \right)\]

\[y = \cos^{2}x:\]

\[\frac{y}{1 - y} = 9 - 6y + 6(1 - y)\]

\[\frac{y}{1 - y} = 9 - 6y + 6 - 6y\]

\[y = (1 - y)(15 - 12y)\]

\[y = 15 - 12y - 15y + 12y^{2}\]

\[12y^{2} - 28y + 15 = 0\]

\[D = 784 - 720 = 64\]

\[y_{1} = \frac{28 - 8}{2 \bullet 12} = \frac{5}{6};\]

\[y_{2} = \frac{28 + 8}{2 \bullet 12} = \frac{3}{2}.\]

\[1)\ \cos^{2}x = \frac{5}{6}\]

\[\frac{1 + \cos{2x}}{2} = \frac{5}{6}\]

\[3 + 3\cos{2x} = 5\]

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

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

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

\[x = \pm \frac{1}{2}\arccos\frac{2}{3} + \pi n.\]

\[2)\ \cos^{2}x = \frac{3}{2}\]

\[x \in \varnothing.\]

\[Область\ определения:\]

\[ctg\ x > 0.\]

\[Ответ:\ \ \frac{1}{2}\arccos\frac{2}{3} + \pi n.\]

\[2)\log_{27}\left( \sin{2x} - \frac{1}{3}\cos x \right) =\]

\[= \frac{1}{3} + \log_{3}\left( - \cos x \right)\ \]

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

\[= \frac{1}{3} + \log_{3}\left( - \cos x \right)\]

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

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

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

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

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

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

\[6\sin x - 1 = - 9\cos^{2}x\]

\[6\sin x - 1 = - 9\left( 1 - \sin^{2}x \right)\]

\[y = \sin x:\]

\[6y - 1 = - 9\left( 1 - y^{2} \right)\]

\[6y - 1 = 9y^{2} - 9\]

\[9y^{2} - 6y - 8 = 0\]

\[D = 36 + 288 = 324\]

\[y_{1} = \frac{6 - 18}{2 \bullet 9} = - \frac{2}{3};\]

\[y_{2} = \frac{6 + 18}{2 \bullet 9} = \frac{4}{3}.\]

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

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

\[2)\ \sin x = \frac{4}{3}\]

\[x \in \varnothing.\]

\[Область\ определения:\]

\[\cos x < 0.\]

\[Ответ:\ \arcsin\frac{2}{3} + (2n + 1)\mathbf{\pi}.\]