Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1368

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

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

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

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

\[\frac{3x}{2} = \arccos 0 + \pi n = \frac{\pi}{2} + \pi n\]

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

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

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

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

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

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

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

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

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

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

\[x = \frac{\text{πn}}{2}.\]

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

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

\[x = \frac{\text{πn}}{3}.\]

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

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

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

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

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

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

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

\[x = \frac{\text{πn}}{2}.\]

\[\cos x - 1 = 0\]

\[\cos x = 1\]

\[x = \arccos 1 + 2\pi n = 2\pi n.\]

\[Ответ:\ \ \frac{\text{πn}}{2}.\]

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

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

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

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

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

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

\[x = \frac{\text{πn}}{2}.\]

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

\[2\cos x = - 1\]

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

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

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

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

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