Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 871

\[1)\sin x + \sin{5x} = \sin{3x}\]

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

\[\sin{3x} \bullet \left( 2\cos{2x} - 1 \right) = 0\]

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

\[3x = \pi n\]

\[x = \frac{\text{πn}}{3}.\]

\[2)\ 2\cos{2x} - 1 = 0\]

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

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

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

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

\[x = \frac{1}{2} \bullet \left( \pm \frac{\pi}{3} + 2\pi n \right)\]

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

\[Ответ:\ \ \frac{\text{πn}}{3};\ \pm \frac{\pi}{6} + \pi n.\]

\[2)\cos{7x} - \cos{3x} = 3\sin{5x}\]

\[- 2 \bullet \sin{5x} \bullet \sin{2x} - 3\sin{5x} = 0\]

\[- \sin{5x} \bullet \left( 2\sin{2x} + 3 \right) = 0\]

\[1)\ \sin{5x} = 0\]

\[5x = \pi n\]

\[x = \frac{\text{πn}}{5}.\]

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

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

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

\[x \in \varnothing.\]

\[Ответ:\ \ \frac{\text{πn}}{5}.\]