Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 127

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

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

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

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

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

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

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

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

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

\[\sin\frac{3x}{2} \cdot \left( \cos\left( \frac{\pi}{2} - \frac{x}{2} \right) - \cos\frac{3x}{2} \right) = 0\]

\[\sin\frac{3x}{2} \cdot ( - 2) \cdot \sin\left( \frac{\pi}{4} + \frac{x}{2} \right) \cdot \sin\left( \frac{\pi}{4} - x \right) = 0\]

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

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

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

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

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

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

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

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

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

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