ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 637

\[\boxed{\mathbf{637}\mathbf{.}}\]

\[3\sin{3x} + 2 \bullet \sin{3x} \bullet \cos{2x} = 0\]

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

\[1)\ \sin{3x} = 0\]

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

\[x = \frac{1}{3} \bullet \pi n = \frac{\text{πn}}{3}.\]

\[2)\ 3 + 2\cos{2x} = 0\]

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

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

\[корней\ нет.\]

\[Ответ:\ \ \frac{\text{πn}}{3}.\]

\[2)\ 6\cos{2x} \bullet \sin x + 7\sin{2x} = 0\]

\[6\cos{2x} \bullet \sin x + 14\sin x \bullet \cos x = 0\]

\[2\sin x \bullet \left( 3\cos{2x} + 7\cos x \right) = 0\]

\[3\cos{2x} + 7\cos x = 0\]

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

\[3\cos^{2}x - 3\left( 1 - \cos^{2}x \right) + 7\cos x = 0\]

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

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

\[y = \cos x:\]

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

\[D = 49 + 72 = 121\]

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

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

\[1)\ 2\sin x = 0\]

\[\sin x = 0\]

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

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

\[корней\ нет.\]

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

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

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