\textbf{Решение:}
а) Преобразуем уравнение, используя определение логарифма:
log 9 ( 3 2 x + 5 2 sin x − 6 cos 2 x − 2 ) = log 9 9 x , \log_9(3^{2x} + 5\sqrt{2}\sin x - 6\cos^2 x - 2) = \log_9 9^x, log 9 ( 3 2 x + 5 2 sin x − 6 cos 2 x − 2 ) = log 9 9 x , 3 2 x + 5 2 sin x − 6 cos 2 x − 2 = 9 x . 3^{2x} + 5\sqrt{2}\sin x - 6\cos^2 x - 2 = 9^x. 3 2 x + 5 2 sin x − 6 cos 2 x − 2 = 9 x .
Так как 3 2 x = 9 x 3^{2x} = 9^x 3 2 x = 9 x 9 x + 5 2 sin x − 6 cos 2 x − 2 = 9 x , 9^x + 5\sqrt{2}\sin x - 6\cos^2 x - 2 = 9^x, 9 x + 5 2 sin x − 6 cos 2 x − 2 = 9 x , 5 2 sin x − 6 cos 2 x − 2 = 0. 5\sqrt{2}\sin x - 6\cos^2 x - 2 = 0. 5 2 sin x − 6 cos 2 x − 2 = 0.
Выразим cos 2 x \cos^2 x cos 2 x sin x \sin x sin x 5 2 sin x − 6 ( 1 − sin 2 x ) − 2 = 0 , 5\sqrt{2}\sin x - 6(1 - \sin^2 x) - 2 = 0, 5 2 sin x − 6 ( 1 − sin 2 x ) − 2 = 0 , 6 sin 2 x + 5 2 sin x − 8 = 0. 6\sin^2 x + 5\sqrt{2}\sin x - 8 = 0. 6 sin 2 x + 5 2 sin x − 8 = 0.
Сделаем замену t = sin x t = \sin x t = sin x 6 t 2 + 5 2 t − 8 = 0. 6t^2 + 5\sqrt{2}t - 8 = 0. 6 t 2 + 5 2 t − 8 = 0.
Найдём дискриминант:
D = ( 5 2 ) 2 − 4 ⋅ 6 ⋅ ( − 8 ) = 50 + 192 = 242 = ( 11 2 ) 2 . D = (5\sqrt{2})^2 - 4\cdot 6 \cdot (-8) = 50 + 192 = 242 = (11\sqrt{2})^2. D = ( 5 2 ) 2 − 4 ⋅ 6 ⋅ ( − 8 ) = 50 + 192 = 242 = ( 11 2 ) 2 .
Тогда
t = − 5 2 ± 11 2 12 . t = \frac{-5\sqrt{2} \pm 11\sqrt{2}}{12}. t = 12 − 5 2 ± 11 2 .
Получаем:
t 1 = 6 2 12 = 2 2 , t 2 = − 16 2 12 = − 4 2 3 . t_1 = \frac{6\sqrt{2}}{12} = \frac{\sqrt{2}}{2}, \qquad
t_2 = \frac{-16\sqrt{2}}{12} = -\frac{4\sqrt{2}}{3}. t 1 = 12 6 2 = 2 2 , t 2 = 12 − 16 2 = − 3 4 2 .
Так как sin x ∈ [ − 1 ; 1 ] \sin x \in [-1; 1] sin x ∈ [ − 1 ; 1 ] t 2 = − 4 2 3 < − 1 t_2 = -\frac{4\sqrt{2}}{3} < -1 t 2 = − 3 4 2 < − 1 sin x = 2 2 . \sin x = \frac{\sqrt{2}}{2}. sin x = 2 2 .
Следовательно,
x = π 4 + 2 π k , x = 3 π 4 + 2 π k , k ∈ Z . x = \frac{\pi}{4} + 2\pi k, \qquad x = \frac{3\pi}{4} + 2\pi k, \quad k \in \mathbb{Z}. x = 4 π + 2 π k , x = 4 3 π + 2 π k , k ∈ Z .
б) Найдём корни на отрезке [ − 2 π ; − π 2 ] \left[-2\pi; -\frac{\pi}{2}\right] [ − 2 π ; − 2 π ] x = π 4 + 2 π k x = \frac{\pi}{4} + 2\pi k x = 4 π + 2 π k − 2 π ≤ π 4 + 2 π k ≤ − π 2 . -2\pi \le \frac{\pi}{4} + 2\pi k \le -\frac{\pi}{2}. − 2 π ≤ 4 π + 2 π k ≤ − 2 π . π \pi π − 2 ≤ 1 4 + 2 k ≤ − 1 2 . -2 \le \frac{1}{4} + 2k \le -\frac{1}{2}. − 2 ≤ 4 1 + 2 k ≤ − 2 1 . 1 4 \frac{1}{4} 4 1 − 9 4 ≤ 2 k ≤ − 3 4 . -\frac{9}{4} \le 2k \le -\frac{3}{4}. − 4 9 ≤ 2 k ≤ − 4 3 . − 9 8 ≤ k ≤ − 3 8 . -\frac{9}{8} \le k \le -\frac{3}{8}. − 8 9 ≤ k ≤ − 8 3 . k k k k = − 1 k = -1 k = − 1 x = π 4 − 2 π = − 7 π 4 . x = \frac{\pi}{4} - 2\pi = -\frac{7\pi}{4}. x = 4 π − 2 π = − 4 7 π .
Для x = 3 π 4 + 2 π k x = \frac{3\pi}{4} + 2\pi k x = 4 3 π + 2 π k − 2 π ≤ 3 π 4 + 2 π k ≤ − π 2 . -2\pi \le \frac{3\pi}{4} + 2\pi k \le -\frac{\pi}{2}. − 2 π ≤ 4 3 π + 2 π k ≤ − 2 π . π \pi π − 2 ≤ 3 4 + 2 k ≤ − 1 2 . -2 \le \frac{3}{4} + 2k \le -\frac{1}{2}. − 2 ≤ 4 3 + 2 k ≤ − 2 1 . 3 4 \frac{3}{4} 4 3 − 11 4 ≤ 2 k ≤ − 5 4 . -\frac{11}{4} \le 2k \le -\frac{5}{4}. − 4 11 ≤ 2 k ≤ − 4 5 . − 11 8 ≤ k ≤ − 5 8 . -\frac{11}{8} \le k \le -\frac{5}{8}. − 8 11 ≤ k ≤ − 8 5 . k k k k = − 1 k = -1 k = − 1 x = 3 π 4 − 2 π = − 5 π 4 . x = \frac{3\pi}{4} - 2\pi = -\frac{5\pi}{4}. x = 4 3 π − 2 π = − 4 5 π .
Отметим корни на тригонометрической окружности.
− 7 π 4 , − 5 π 4 . -\frac{7\pi}{4}, \quad -\frac{5\pi}{4}. − 4 7 π , − 4 5 π .
Ответ:
а) x = π 4 + 2 π k , x = 3 π 4 + 2 π k , k ∈ Z ; x = \frac{\pi}{4} + 2\pi k,\; x = \frac{3\pi}{4} + 2\pi k,\; k \in \mathbb{Z}; x = 4 π + 2 π k , x = 4 3 π + 2 π k , k ∈ Z ;
б) − 7 π 4 ; − 5 π 4 -\frac{7\pi}{4};\, -\frac{5\pi}{4} − 4 7 π ; − 4 5 π
Следовательно, корни, принадлежащие отрезку: