Страница 155 номер 523, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Решить уравнение: 1) \(1 - \cos x = 2 \sin\frac{x}{2}\); 2) \(1 + \cos x = 2 \cos\frac{x}{2}\); 3) \(1 + \cos\frac{x}{2} = 2 \sin\left(\frac{x}{4} - \frac{3\pi}{2}\right)\); 4) \(1 + \cos 8x = 2 \cos 4x\); 5) \(2 \sin^2 \frac{x}{2} + \frac{1}{2} \sin 2x = 1\); 6) \(2 \cos^2 x - \frac{1}{2} \sin 4x = 1\).
1) \(1 - \cos x = 2 \sin\frac{x}{2}\)
\(\cos^{2} \dfrac{x}{2} + \sin^{2} \dfrac{x}{2} - \cos^{2} \dfrac{x}{2} + \sin^{2} \dfrac{x}{2} - 2 \sin\dfrac{x}{2} = 0,\)
\(2 \sin\dfrac{x}{2}\left(\sin\dfrac{x}{2} - 1\right) = 0,\)
\(\sin\frac{x}{2} = 0\) или \(\sin\frac{x}{2} = 1\)
\(\frac{x}{2} = \pi k\) или \(\frac{x}{2} = \frac{\pi}{2} + 2\pi k\)
\(x_{1} = 2\pi k, x_{2} = \pi + 4\pi k, k \in Z;\)
2) \(1 + \cos x = 2 \cos\frac{x}{2}\)
\(\cos^{2} \dfrac{x}{2} + \sin^{2} \dfrac{x}{2} + \cos^{2} \dfrac{x}{2} - \sin^{2} \dfrac{x}{2} - 2 \cos\dfrac{x}{2} = 0,\)
\(2 \cos\dfrac{x}{2}\left(\cos\dfrac{x}{2} - 1\right) = 0,\)
\(\cos\frac{x}{2} = 0\) или \(\cos\frac{x}{2} = 1\)
\(\frac{x}{2} = \frac{\pi}{2} + \pi k\) или \(\frac{x}{2} = 2\pi k\)
\(x_{1} = \pi + 2\pi k, x_{2} = 4\pi k, k \in Z;\)
3) \(1 + \cos\frac{x}{2} = 2 \sin\left(\frac{x}{4} - \frac{3\pi}{2}\right)\)
\(\cos^{2} \dfrac{x}{4} + \sin^{2} \dfrac{x}{4} + \cos^{2} \dfrac{x}{4} - \sin^{2} \dfrac{x}{4} =\)
\(= 2 \sin\dfrac{x}{4}\cos\dfrac{3\pi}{2} - 2 \cos\dfrac{x}{4}\sin\dfrac{3\pi}{2},\)
\(2 \cos^{2} \dfrac{x}{4} - 2 \cos\dfrac{x}{4} = 0,\)
\(2 \cos\dfrac{x}{4}\left(\cos\dfrac{x}{4} - 1\right) = 0,\)
\(\cos\frac{x}{4} = 0\) или \(\cos\frac{x}{4} = 1\)
\(\frac{x}{4} = \frac{\pi}{2} + \pi k\) или \(\frac{x}{4} = 2\pi k\)
\(x_{1} = 2\pi + 4\pi k, x_{2} = 8\pi k, k \in Z;\)
4) \(1 + \cos 8x = 2 \cos 4x\)
\(\cos^{2} 4x + \sin^{2} 4x + \cos^{2} 4x - \sin^{2} 4x - 2 \cos 4x = 0,\)
\(2 \cos 4x (\cos 4x - 1) = 0,\)
\(\cos 4x = 0\) или \(\cos 4x = 1\)
\(4x = \frac{\pi}{2} + \pi k\) или \(4x = 2\pi k\)
\(x_{1} = \dfrac{\pi}{8} + \dfrac{\pi}{4} k, x_{2} = \dfrac{\pi}{2} k, k \in Z;\)
5) \(2 \sin^{2} \frac{x}{2} + \frac{1}{2} \sin 2x = 1\)
\(1 - \cos x + \sin x \cos x = 1,\)
\(\cos x (\sin x - 1) = 0,\)
\(\cos x = 0\) или \(\sin x = 1\)
\(x_{1} = \dfrac{\pi}{2} + \pi k, x_{2} = \dfrac{\pi}{2} + 2\pi k, k \in Z.\)
Ответ: \(x = \frac{\pi}{2} + \pi k, k \in Z\).
6) \(2 \cos^{2} x - \frac{1}{2} \sin 4x = 1\)
\(1 + \cos 2x - \sin 2x \cos 2x = 1,\)
\(\cos 2x (1 - \sin 2x) = 0,\)
\(\cos 2x = 0\) или \(\sin 2x = 1\)
\(2x = \frac{\pi}{2} + \pi k\) или \(2x = \frac{\pi}{2} + 2\pi k\)
\(x = \dfrac{\pi}{4} + \dfrac{\pi}{2} k, k \in Z.\)
