Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 684

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\[\left| \cos x \right| - \cos{3x} = \sin{2x}\]

\[1)\ \cos x < 0:\]

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

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

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

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

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

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

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

\[y = \sin x:\]

\[y + 1 - 2y^{2} = 0\]

\[2y^{2} - y - 1 = 0\]

\[D = 1 + 8 = 9\]

\[y_{1} = \frac{1 - 3}{2 \bullet 2} = - \frac{2}{4} = - \frac{1}{2};\]

\[y_{2} = \frac{1 + 3}{2 \bullet 2} = 1.\]

\[\cos x = 0\]

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

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

\[x = ( - 1)^{n + 1} \bullet \arcsin\frac{1}{2} + \pi n\]

\[x = ( - 1)^{n + 1} \bullet \frac{\pi}{6} + \pi n.\]

\[\sin x = 1\]

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

\[2)\ \cos x > 0:\]

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

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

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

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

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

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

\[x = \frac{\text{πn}}{2}.\]

\[- \frac{\pi}{2} + \pi n < x < \frac{\pi}{2} + \pi n:\]

\[x = 2\pi n.\]

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

\[2\sin x = 1\]

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

\[x = ( - 1)^{n} \bullet \arcsin\frac{1}{2} + \pi n\]

\[x = ( - 1)^{n} \bullet \frac{\pi}{6} + \pi n.\]

\[Ответ:\ \ \]

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

\[x = \frac{\pi}{6} + \pi n.\]