ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1566

\[\boxed{\mathbf{1566}\mathbf{.}}\]

\[\log_{2}\left( 4\cos x + 3 \right) \bullet \log_{6}\left( 4\cos x + 3 \right) =\]

\[= \log_{2}\left( 4\cos x + 3 \right) + {\text{lo}g_{6}}\left( 4\cos x + 3 \right)\]

\[y = 4\cos x + 3:\]

\[\log_{2}y \bullet \log_{6}y = \log_{2}y + {\text{lo}g_{6}}y\]

\[\log_{2}y \bullet \log_{6}y - \log_{2}y - \log_{6}y = 0\]

\[\log_{2}y \bullet \left( \log_{6}y - 1 \right) - \log_{6}y = 0\]

\[\log_{2}y \bullet \left( \log_{6}y - \log_{6}6 \right) - \log_{6}y = 0\]

\[\frac{\log_{6}y}{\log_{6}2} \bullet \log_{6}\frac{y}{6} - \log_{6}y = 0\]

\[\log_{6}y \bullet \log_{6}\frac{y}{6} - \log_{6}2 \bullet \log_{6}y = 0\]

\[\log_{6}y \bullet \left( \log_{6}\frac{y}{6} - \log_{6}2 \right) = 0\]

\[\log_{6}y \bullet \log_{6}\frac{y}{12} = 0.\]

\[1)\ \log_{6}y = 0\]

\[y = 1\]

\[4\cos x + 3 = 1\]

\[4\cos x = - 2\]

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

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

\[= \pm \left( \pi - \frac{\pi}{3} \right) + 2\pi n =\]

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

\[2)\ \log_{6}\frac{y}{12} = 0\]

\[\frac{y}{12} = 1\]

\[y = 12\]

\[4\cos x + 3 = 12\]

\[4\cos x = 9\]

\[\cos x = \frac{9}{4}\]

\[корней\ нет.\]

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