Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1604

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\[На\ интервале\ \left( - \frac{\pi}{6};\ \frac{\pi}{2} \right):\]

\[\cos\left( 5x + \frac{\pi}{2} \right) + 2\sin x \bullet \cos{2x} = 0\]

\[- \sin{5x} + \sin(x + 2x) + \sin(x - 2x) = 0\]

\[- \sin{5x} + \sin{3x} - \sin x = 0\]

\[\sin{3x} - \left( \sin{5x} + \sin x \right) = 0\]

\[\sin{3x} - 2 \bullet \sin\frac{5x + x}{2} \bullet \cos\frac{5x - x}{2} = 0\]

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

\[\sin{3x} \bullet \left( 1 - 2\cos{2x} \right) = 0\]

\[1)\ \sin{3x} = 0\]

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

\[x = \frac{\text{πn}}{3}.\]

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

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

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

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

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

\[x = \frac{1}{2} \bullet \left( \pm \frac{\pi}{3} + 2\pi n \right)\]

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

\[x_{1} = - \frac{\pi}{6};\text{\ \ }x_{2} = \frac{\pi}{6};\]

\[x_{3} = 0;\text{\ \ }x_{4} = \frac{\pi}{3}.\]

\[Ответ:\ \ x = \frac{\pi}{3}.\]