Задание 13. Математика. ЕГЭ 2026. Ященко. Вариант 9

а) $$\cos 2x \cdot \sin 4x - \cos 4x \cdot \sin \frac{5 \pi}{6} = \cos \left( 2x - \frac{\pi}{2} \right) $$ $$2 \cos^2 2x \cdot \sin 2x - \frac{1}{2} \cos 4x - \sin 2x = 0$$ $$\sin 2x \cdot (2 \cos^2 2x - 1) - \frac{1}{2} \cos 4x = 0$$ $$\cos 4x \cdot \left( \sin 2x - \frac{1}{2} \right) = 0$$ Значит, либо \(\cos 4x = 0 \, \Rightarrow \) \(4x = \cfrac{\pi}{2} + \pi m, \, m \in \mathbb{Z}, \, \Rightarrow \) \(x = \cfrac{\pi}{8} + \cfrac{\pi m}{4}, \, m \in \mathbb{Z}.\)

Либо \(\sin 2x = \cfrac{1}{2} \, \Rightarrow \) \(2x = \cfrac{\pi}{6} + 2 \pi n, \, n \in \mathbb{Z};\) \(2x = \cfrac{5\pi}{6} + 2 \pi k, \, k \in \mathbb{Z} \, \Rightarrow\) \(x = \cfrac{\pi}{12} + \pi n, \, n \in \mathbb{Z};\) \(x = \cfrac{5\pi}{12} + \pi k, \, k \in \mathbb{Z}.\)

б) Отбор корней. I $$- 3 \pi \leqslant \frac{\pi}{8} + \frac{\pi m}{4} \leqslant -2 \pi$$ $$-24 \leqslant 1 + 2m \leqslant - 16$$ $$-25 \leqslant 2m \leqslant -17$$ Получаем, что \(m = -9, \, -10, \, -11, \, -12 \, \Rightarrow\) \(x = - \cfrac{17 \pi}{8}, \, - \cfrac{19 \pi}{8}, \, - \cfrac{21 \pi}{8}, \, - \cfrac{23 \pi}{8}.\)

II. $$- 3 \pi \leqslant \frac{\pi}{12} + \pi n \leqslant -2 \pi$$ $$-36 \leqslant 1 + 12 n \leqslant -24$$ $$-37 \leqslant 12 n \leqslant -25$$ Значит \(n = -3 \, \Rightarrow \) \(x = - \cfrac{35 \pi}{12}.\)

III. $$- 3 \pi \leqslant \frac{5 \pi}{12} + \pi k \leqslant -2 \pi$$ $$-36 \leqslant 5 + 12 k \leqslant -24$$ $$-41 \leqslant 12 k \leqslant -29$$ Значит \(k = -3 \, \Rightarrow \) \(x = - \cfrac{31 \pi}{12}.\)

Ответ: а) \(\cfrac{\pi}{8} + \cfrac{\pi m}{4} \, m \in \mathbb{Z};\) \(\cfrac{\pi}{12} + \pi n, \, n \in \mathbb{Z};\) \(\cfrac{5\pi}{12} + \pi k, \, k \in \mathbb{Z};\) б) \(-\cfrac{-35 \pi}{12};\) \(-\cfrac{-23 \pi}{8};\) \(-\cfrac{-21 \pi}{8};\) \(-\cfrac{-31 \pi}{12};\) \(-\cfrac{-19 \pi}{8};\) \(-\cfrac{-17 \pi}{8}.\)