Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 866

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

\[2 \bullet \cos\frac{2x + x}{2} \bullet \cos\frac{2x - x}{2} = 0\]

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

\[1)\ \cos\frac{3x}{2} = 0\]

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

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

\[2)\ \cos\frac{x}{2} = 0\]

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

\[x = \pi + 2\pi n.\]

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

\[2)\cos x - \cos{5x} = 0\]

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

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

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

\[2x = \pi n\]

\[x = \frac{\text{πn}}{2}.\]

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

\[3x = \pi n\]

\[x = \frac{\text{πn}}{3}.\]

\[Ответ:\ \ \frac{\text{πn}}{2};\ \frac{\text{πn}}{3}.\]

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

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

\[\sin{2x} \bullet \cos x = \sin{2x}\]

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

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

\[2x = \pi n\]

\[x = \frac{\text{πn}}{2}.\]

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

\[\cos x = 1\]

\[x = 2\pi n.\]

\[Ответ:\ \ \frac{\text{πn}}{2}.\]

\[4)\sin x + \sin{2x} + \sin{3x} = 0\]

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

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

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

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

\[2x = \pi n\]

\[x = \frac{\text{πn}}{2}.\]

\[2)\ 2\cos x + 1 = 0\]

\[2\cos x = - 1\]

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

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

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

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