Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 770

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

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

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

\[Первое\ уравнение:\]

\[\cos x = 0\]

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

\[Второе\ уравнение:\]

\[\cos x - 1 = 0\]

\[\cos x = 1\]

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

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

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

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

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

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

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

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

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

\[Первое\ уравнение:\]

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

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

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

\[Второе\ уравнение:\]

\[\sin\left( x - \frac{\pi}{4} \right) = 0\]

\[x - \frac{\pi}{4} = \arcsin 0 + \pi n = \pi n\]

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

\[Третье\ уравнение:\]

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

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

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

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

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

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

\[\frac{x}{2} = \frac{3\pi}{4} + \pi n\]

\[x = 2 \bullet \left( \frac{3\pi}{4} + \pi n \right)\]

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

\[Ответ:\ \ \frac{2\pi n}{3}\text{\ \ }\frac{\pi}{4} + \pi n\ \ \frac{3\pi}{2} + 2\pi n.\]