Задание 14. Математика. ЕГЭ 2027. Ященко. Вариант 7

а) $$3^{\sin^2 3x} \left( 1 + 9 \right) = 5^{\sin^2 3x} \left( 1 + 5 \right) \\[1em] \left( \frac{3}{5} \right)^{\sin^2 3x} = \frac{3}{5} \\[1em] \sin^2 3x = 1 \\[2em] \begin{array}{l l | l} \sin 3x = 1 & \quad & \quad \sin 3x = -1 \\[1em] 3x = \cfrac{\pi}{2} + 2 \pi n, \, n \in \mathbb{Z} & \quad & \quad 3x = -\cfrac{\pi}{2} + 2 \pi k, \, k \in \mathbb{Z} \\[1em] x = \cfrac{\pi}{6} + \cfrac{2 \pi n}{3} \, n \in \mathbb{Z} & \quad & \quad x = - \cfrac{\pi}{6} + \cfrac{2 \pi k}{3} \, k \in \mathbb{Z} \end{array}$$

б) Отбор корней $$ \begin{array}{l l | l} 3 \pi \leqslant \cfrac{\pi}{6} + \cfrac{2 \pi n}{3} \leqslant 4 \pi & \quad & \quad 3 \pi \leqslant - \cfrac{\pi}{6} + \cfrac{2 \pi k}{3} \leqslant 4 \pi \\[1em] 18 \leqslant 1 + 4n \leqslant 24 & \quad & \quad 18 \leqslant -1 + 4k \leqslant 24 \\[1em] 17 \leqslant 4n \leqslant 23 & \quad & \quad 19 \leqslant 4k \leqslant 25 \\[1em] n = 5 & \quad & \quad k = 5, \, 6 \\[1em] x = \cfrac{\pi}{6} + \cfrac{10 \pi}{3} = \cfrac{7 \pi}{2} & \quad & \quad x = - \cfrac{\pi}{6} + \cfrac{10 \pi}{3} = \cfrac{19 \pi}{6} \\[1em] & \quad & \quad x = - \cfrac{\pi}{6} + 4 \pi = \cfrac{23 \pi}{6} \end{array} $$

Ответ: а) \( \cfrac{\pi}{6} + \cfrac{2 \pi n}{3} \, n \in \mathbb{Z};\) \( - \cfrac{\pi}{6} + \cfrac{2 \pi k}{3} \, k \in \mathbb{Z};\) б) \(\cfrac{19 \pi}{6}; \, \cfrac{7 \pi}{2}; \, \cfrac{23 \pi}{6}. \)