a) Упростим уравнение, используя формулу приведения:
sin ( 2 x − 3 π 2 ) = cos 2 x \sin \left( 2x - \frac{3\pi}{2} \right) = \cos 2x sin ( 2 x − 2 3 π ) = cos 2 x
Так как cos 2 π 3 = − 1 2 \cos \frac{2\pi}{3} = -\frac{1}{2} cos 3 2 π = − 2 1 sin 2 x ⋅ sin 4 x − 1 2 cos 4 x = cos 2 x \sin 2x \cdot \sin 4x - \frac{1}{2} \cos 4x = \cos 2x sin 2 x ⋅ sin 4 x − 2 1 cos 4 x = cos 2 x
Используем формулу sin 4 x = 2 sin 2 x cos 2 x \sin 4x = 2\sin 2x \cos 2x sin 4 x = 2 sin 2 x cos 2 x 2 sin 2 2 x cos 2 x − 1 2 cos 4 x = cos 2 x 2\sin^2 2x \cos 2x - \frac{1}{2} \cos 4x = \cos 2x 2 sin 2 2 x cos 2 x − 2 1 cos 4 x = cos 2 x
Используем формулу cos 4 x = 1 − 2 sin 2 2 x \cos 4x = 1 - 2\sin^2 2x cos 4 x = 1 − 2 sin 2 2 x 2 sin 2 2 x cos 2 x − 1 2 ( 1 − 2 sin 2 2 x ) = cos 2 x 2\sin^2 2x \cos 2x - \frac{1}{2}(1 - 2\sin^2 2x) = \cos 2x 2 sin 2 2 x cos 2 x − 2 1 ( 1 − 2 sin 2 2 x ) = cos 2 x 2 sin 2 2 x cos 2 x − 1 2 + sin 2 2 x = cos 2 x 2\sin^2 2x \cos 2x - \frac{1}{2} + \sin^2 2x = \cos 2x 2 sin 2 2 x cos 2 x − 2 1 + sin 2 2 x = cos 2 x
Переносим все в одну сторону:
2 sin 2 2 x cos 2 x + sin 2 2 x − cos 2 x − 1 2 = 0 2\sin^2 2x \cos 2x + \sin^2 2x - \cos 2x - \frac{1}{2} = 0 2 sin 2 2 x cos 2 x + sin 2 2 x − cos 2 x − 2 1 = 0
Группируем:
sin 2 2 x ( 2 cos 2 x + 1 ) − ( cos 2 x + 1 2 ) = 0 \sin^2 2x (2\cos 2x + 1) - (\cos 2x + \frac{1}{2}) = 0 sin 2 2 x ( 2 cos 2 x + 1 ) − ( cos 2 x + 2 1 ) = 0
Замечаем, что cos 2 x + 1 2 = 1 2 ( 2 cos 2 x + 1 ) \cos 2x + \frac{1}{2} = \frac{1}{2}(2\cos 2x + 1) cos 2 x + 2 1 = 2 1 ( 2 cos 2 x + 1 ) ( 2 cos 2 x + 1 ) ( sin 2 2 x − 1 2 ) = 0 (2\cos 2x + 1)(\sin^2 2x - \frac{1}{2}) = 0 ( 2 cos 2 x + 1 ) ( sin 2 2 x − 2 1 ) = 0
Решаем первое уравнение:
2 cos 2 x + 1 = 0 ⇒ cos 2 x = − 1 2 2\cos 2x + 1 = 0 \Rightarrow \cos 2x = -\frac{1}{2} 2 cos 2 x + 1 = 0 ⇒ cos 2 x = − 2 1 2 x = ± 2 π 3 + 2 π k 2x = \pm \frac{2\pi}{3} + 2\pi k 2 x = ± 3 2 π + 2 π k x = ± π 3 + π k , k ∈ Z x = \pm \frac{\pi}{3} + \pi k, \quad k \in \mathbb{Z} x = ± 3 π + π k , k ∈ Z
Решаем второе уравнение:
sin 2 2 x = 1 2 ⇒ sin 2 x = ± 2 2 \sin^2 2x = \frac{1}{2} \Rightarrow \sin 2x = \pm \frac{\sqrt{2}}{2} sin 2 2 x = 2 1 ⇒ sin 2 x = ± 2 2 2 x = π 4 + π 2 k 2x = \frac{\pi}{4} + \frac{\pi}{2}k 2 x = 4 π + 2 π k x = π 8 + π 4 k , k ∈ Z x = \frac{\pi}{8} + \frac{\pi}{4}k, \quad k \in \mathbb{Z} x = 8 π + 4 π k , k ∈ Z
Таким образом получаем:
x = π 8 + π 4 k , k ∈ Z x = \frac{\pi}{8} + \frac{\pi}{4}k, \quad k \in \mathbb{Z} x = 8 π + 4 π k , k ∈ Z x = − π 3 + π k , k ∈ Z x = -\frac{\pi}{3} + \pi k, \quad k \in \mathbb{Z} x = − 3 π + π k , k ∈ Z x = π 3 + π k , k ∈ Z x = \frac{\pi}{3} + \pi k, \quad k \in \mathbb{Z} x = 3 π + π k , k ∈ Z
б) Отберём корни, принадлежащие отрезку [ − 7 π 2 ; − 5 π 2 ] \left[ -\frac{7\pi}{2}; -\frac{5\pi}{2} \right] [ − 2 7 π ; − 2 5 π ]
Для x = π 8 + π 4 k , k ∈ Z x = \frac{\pi}{8} + \frac{\pi}{4}k, \ k \in \mathbb{Z} x = 8 π + 4 π k , k ∈ Z − 7 π 2 ≤ π 8 + π 4 k ≤ − 5 π 2 − 28 π 8 ≤ π 8 + 2 π 8 k ≤ − 20 π 8 − 28 π 8 − π 8 ≤ 2 π 8 k ≤ − 20 π 8 − π 8 − 29 π 8 ≤ π 4 k ≤ − 21 π 8 − 29 2 ≤ k ≤ − 21 2 ⇒ k = − 14 , − 13 , − 12 , − 11 \begin{align*}
-\frac{7\pi}{2} &\le \frac{\pi}{8} + \frac{\pi}{4}k \le -\frac{5\pi}{2} \\
-\frac{28\pi}{8} &\le \frac{\pi}{8} + \frac{2\pi}{8}k \le -\frac{20\pi}{8} \\
-\frac{28\pi}{8} - \frac{\pi}{8} &\le \frac{2\pi}{8}k \le -\frac{20\pi}{8} - \frac{\pi}{8} \\
-\frac{29\pi}{8} &\le \frac{\pi}{4}k \le -\frac{21\pi}{8} \\
-\frac{29}{2} &\le k \le -\frac{21}{2} \quad \Rightarrow \quad k = -14, -13, -12, -11
\end{align*} − 2 7 π − 8 28 π − 8 28 π − 8 π − 8 29 π − 2 29 ≤ 8 π + 4 π k ≤ − 2 5 π ≤ 8 π + 8 2 π k ≤ − 8 20 π ≤ 8 2 π k ≤ − 8 20 π − 8 π ≤ 4 π k ≤ − 8 21 π ≤ k ≤ − 2 21 ⇒ k = − 14 , − 13 , − 12 , − 11 k = − 14 k = -14 k = − 14 x = π 8 − 14 π 4 = π 8 − 28 π 8 = − 27 π 8 x = \frac{\pi}{8} - \frac{14\pi}{4} = \frac{\pi}{8} - \frac{28\pi}{8} = -\frac{27\pi}{8} x = 8 π − 4 14 π = 8 π − 8 28 π = − 8 27 π k = − 13 k = -13 k = − 13 x = π 8 − 13 π 4 = π 8 − 26 π 8 = − 25 π 8 x = \frac{\pi}{8} - \frac{13\pi}{4} = \frac{\pi}{8} - \frac{26\pi}{8} = -\frac{25\pi}{8} x = 8 π − 4 13 π = 8 π − 8 26 π = − 8 25 π k = − 12 k = -12 k = − 12 x = π 8 − 12 π 4 = π 8 − 24 π 8 = − 23 π 8 x = \frac{\pi}{8} - \frac{12\pi}{4} = \frac{\pi}{8} - \frac{24\pi}{8} = -\frac{23\pi}{8} x = 8 π − 4 12 π = 8 π − 8 24 π = − 8 23 π k = − 11 k = -11 k = − 11 x = π 8 − 11 π 4 = π 8 − 22 π 8 = − 21 π 8 x = \frac{\pi}{8} - \frac{11\pi}{4} = \frac{\pi}{8} - \frac{22\pi}{8} = -\frac{21\pi}{8} x = 8 π − 4 11 π = 8 π − 8 22 π = − 8 21 π Для x = π 3 + π n , n ∈ Z x = \frac{\pi}{3} + \pi n, \ n \in \mathbb{Z} x = 3 π + π n , n ∈ Z − 7 π 2 ≤ π 3 + π k ≤ − 5 π 2 − 21 π 6 ≤ 2 π 6 + 6 π k 6 ≤ − 15 π 6 − 21 π 6 − 2 π 6 ≤ 6 π k 6 ≤ − 15 π 6 − 2 π 6 − 23 π 6 ≤ π k ≤ − 17 π 6 − 23 6 ≤ k ≤ − 17 6 ⇒ k = − 3 \begin{align*}
-\frac{7\pi}{2} &\le \frac{\pi}{3} + \pi k \le -\frac{5\pi}{2} \\
-\frac{21\pi}{6} &\le \frac{2\pi}{6} + \frac{6\pi k}{6} \le -\frac{15\pi}{6} \\
-\frac{21\pi}{6} - \frac{2\pi}{6} &\le \frac{6\pi k}{6} \le -\frac{15\pi}{6} - \frac{2\pi}{6} \\
-\frac{23\pi}{6} &\le \pi k \le -\frac{17\pi}{6} \\
-\frac{23}{6} &\le k \le -\frac{17}{6} \quad \Rightarrow \quad k = -3
\end{align*} − 2 7 π − 6 21 π − 6 21 π − 6 2 π − 6 23 π − 6 23 ≤ 3 π + π k ≤ − 2 5 π ≤ 6 2 π + 6 6 π k ≤ − 6 15 π ≤ 6 6 π k ≤ − 6 15 π − 6 2 π ≤ π k ≤ − 6 17 π ≤ k ≤ − 6 17 ⇒ k = − 3 k = − 3 : x = π 3 − 3 π = − 8 π 3 k = -3: \quad x = \frac{\pi}{3} - 3\pi = -\frac{8\pi}{3} k = − 3 : x = 3 π − 3 π = − 3 8 π Для x = − π 3 + π k , k ∈ Z x = -\frac{\pi}{3} + \pi k, \ k \in \mathbb{Z} x = − 3 π + π k , k ∈ Z − 7 π 2 ≤ − π 3 + π k ≤ − 5 π 2 − 21 π 6 ≤ − 2 π 6 + 6 π k 6 ≤ − 15 π 6 − 21 π 6 + 2 π 6 ≤ 6 π k 6 ≤ − 15 π 6 + 2 π 6 − 19 π 6 ≤ π k ≤ − 13 π 6 − 19 6 ≤ k ≤ − 13 6 ⇒ k = − 3 , − 2 \begin{align*}
-\frac{7\pi}{2} &\le -\frac{\pi}{3} + \pi k \le -\frac{5\pi}{2} \\
-\frac{21\pi}{6} &\le -\frac{2\pi}{6} + \frac{6\pi k}{6} \le -\frac{15\pi}{6} \\
-\frac{21\pi}{6} + \frac{2\pi}{6} &\le \frac{6\pi k}{6} \le -\frac{15\pi}{6} + \frac{2\pi}{6} \\
-\frac{19\pi}{6} &\le \pi k \le -\frac{13\pi}{6} \\
-\frac{19}{6} &\le k \le -\frac{13}{6} \quad \Rightarrow \quad k = -3, -2
\end{align*} − 2 7 π − 6 21 π − 6 21 π + 6 2 π − 6 19 π − 6 19 ≤ − 3 π + π k ≤ − 2 5 π ≤ − 6 2 π + 6 6 π k ≤ − 6 15 π ≤ 6 6 π k ≤ − 6 15 π + 6 2 π ≤ π k ≤ − 6 13 π ≤ k ≤ − 6 13 ⇒ k = − 3 , − 2 k = − 3 k = -3 k = − 3 x = − π 3 − 3 π = − 10 π 3 x = -\frac{\pi}{3} - 3\pi = -\frac{10\pi}{3} x = − 3 π − 3 π = − 3 10 π k = − 2 k = -2 k = − 2 x = − π 3 − 2 π = − 7 π 3 x = -\frac{\pi}{3} - 2\pi = -\frac{7\pi}{3} x = − 3 π − 2 π = − 3 7 π − 7 π 3 > − 5 π 2 -\frac{7\pi}{3} > -\frac{5\pi}{2} − 3 7 π > − 2 5 π Корни, принадлежащие отрезку − 27 π 8 , − 10 π 3 , − 25 π 8 , − 23 π 8 , − 8 π 3 , − 21 π 8 {-\frac{27\pi}{8},\ -\frac{10\pi}{3},\ -\frac{25\pi}{8},\ -\frac{23\pi}{8},\ -\frac{8\pi}{3},\ -\frac{21\pi}{8}} − 8 27 π , − 3 10 π , − 8 25 π , − 8 23 π , − 3 8 π , − 8 21 π
Ответ: а) π 8 + π 4 k , k ∈ Z ; − π 3 + π k , k ∈ Z ; π 3 + π k , k ∈ Z ; \frac{\pi}{8} + \frac{\pi}{4}k, \ k \in \mathbb{Z}; \ -\frac{\pi}{3} + \pi k, \ k \in \mathbb{Z}; \ \frac{\pi}{3} + \pi k, \ k \in \mathbb{Z}; 8 π + 4 π k , k ∈ Z ; − 3 π + π k , k ∈ Z ; 3 π + π k , k ∈ Z ;
б) − 27 π 8 , − 10 π 3 , − 25 π 8 , − 23 π 8 , − 8 π 3 , − 21 π 8 -\frac{27\pi}{8},\ -\frac{10\pi}{3},\ -\frac{25\pi}{8},\ -\frac{23\pi}{8},\ -\frac{8\pi}{3},\ -\frac{21\pi}{8} − 8 27 π , − 3 10 π , − 8 25 π , − 8 23 π , − 3 8 π , − 8 21 π