Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 675

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\[1)\sin x + \sin{2x} + \sin{3x} = 0\]

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

\[2 \bullet \sin\frac{4x}{2} \bullet \cos\left( - \frac{2x}{2} \right) + \sin{2x} = 0\]

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

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

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

\[2x = \arcsin 0 + \pi n = \pi n\]

\[x = \frac{1}{2} \bullet \pi n = \frac{\text{πn}}{2}.\]

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

\[2\cos x = - 1\]

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

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

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

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

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

\[2)\cos x - \cos{3x} = \cos{2x} - \cos{4x}\]

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

\[= - 2 \bullet \sin\frac{2x + 4x}{2} \bullet \sin\frac{2x - 4x}{2}\]

\[\sin\frac{4x}{2} \bullet \sin\left( - \frac{2x}{2} \right) =\]

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

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

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

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

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

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

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

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

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

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

\[\frac{x}{2} = \arcsin 0 + \pi n = \pi n\]

\[x = 2\pi n.\]

\[3)\ \cos\frac{5x}{2} = 0\]

\[\frac{5x}{2} = \arccos 0 + \pi n = \frac{\pi}{2} + \pi n\]

\[x = \frac{2}{5} \bullet \left( \frac{\pi}{2} + \pi n \right) = \frac{\pi}{5} + \frac{2\pi n}{5}.\]

\[Ответ:\ \ \pi n;\ \ \frac{\pi}{5} + \frac{2\pi n}{5}.\]