Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 523

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

\[2\sin^{2}\frac{x}{2} - 2\sin\frac{x}{2} = 0\]

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

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

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

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

\[x = 2\pi n\]

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

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

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

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

\[x = \pi + 4\pi n\ \]

\[Ответ:\ \ 2\pi n;\ \ \pi + 4\pi n.\]

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

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

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

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

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

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

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

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

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

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

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

\[x = 4\pi n\]

\[Ответ:\ \ \pi + 2\pi n;\ \ 4\pi n.\]

\[3)\ 1 + \cos\frac{x}{2} = 2\sin\left( \frac{x}{4} - \frac{3\pi}{2} \right)\]

\[1 + \cos\frac{x}{2} = - 2\sin\left( \frac{3\pi}{2} - \frac{x}{4} \right)\]

\[1 + \cos\frac{x}{2} = 2\cos\frac{x}{4}\]

\[2\cos^{2}\frac{x}{4} - 2\cos\frac{x}{4} = 0\]

\[2\cos\frac{x}{4}\left( \cos\frac{x}{4} - 1 \right) = 0\]

\[1)\ 2\cos\frac{x}{4} = 0\]

\[\cos\frac{x}{4} = 0\]

\[\frac{x}{4} = \arccos 0 + \pi n = \frac{\pi}{2} + \pi n\]

\[x = 4 \bullet \left( \frac{\pi}{2} + \pi n \right)\]

\[x = 2\pi + 4\pi n\]

\[2)\ \cos\frac{x}{4} - 1 = 0\]

\[\cos\frac{x}{4} = 1\]

\[\frac{x}{4} = \pm \arccos 1 + 2\pi n = 2\pi n\]

\[x = 8\pi n\]

\[Ответ:\ \ 2\pi + 4\pi n;\ \ 8\pi n.\]

\[4)\ 1 + \cos{8x} = 2\cos{4x}\]

\[2\cos^{2}{4x} - 2\cos{4x} = 0\]

\[2\cos{4x}\left( \cos{4x} - 1 \right) = 0\]

\[1)\ 2\cos{4x} = 0\]

\[\cos{4x} = 0\]

\[4x = \arccos 0 + \pi n = \frac{\pi}{2} + \pi n\]

\[x = \frac{\pi}{8} + \frac{\text{πn}}{4}\]

\[2)\ \cos{4x} - 1 = 0\]

\[\cos{4x} = 1\]

\[4x = \pm \arccos 1 + 2\pi n = 2\pi n\]

\[x = \frac{\text{πn}}{2}\]

\[Ответ:\ \ \frac{\pi}{8} + \frac{\text{πn}}{4};\ \ \frac{\text{πn}}{2}.\]

\[5)\ 2\sin^{2}\frac{x}{2} + \frac{1}{2}\sin{2x} = 1\]

\[- \cos x + \sin x \bullet \cos x = 0\]

\[\cos x\left( \sin x - 1 \right) = 0\]

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

\[x = \arccos 0 + \pi n\]

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

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

\[\sin x = 1\]

\[x = \arcsin 1 + 2\pi n\]

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

\[Ответ:\ \ \frac{\pi}{2} + \pi n.\]

\[6)\ 2\cos^{2}x - \frac{1}{2}\sin{4x} = 1\]

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

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

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

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

\[x = \frac{\pi}{4} + \frac{\text{πn}}{2}\]

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

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

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

\[x = \frac{\pi}{4} + \pi n\]

\[Ответ:\ \ \frac{\pi}{4} + \frac{\text{πn}}{2}.\]