Страница 429 номер 1599, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(\cos x + \cos{2x} + \cos{3x} = 0\)
\(\cos{2x} + 2\cos{2x}\cos x = 0,\)
\(\cos{2x}\left( 1 + 2\cos x \right) = 0,\)
\(\cos{2x} = 0\) или \(\cos x = - \frac{1}{2}\)
\(x_{1} = \dfrac{\pi}{4} + \dfrac{\pi n}{2},\)
\(x_{2} = \pm \dfrac{2\pi}{3} + 2\pi n, n \in Z;\)

2) \(\cos^{3}x - 3\cos^{2}x + \cos x + \sin{2x} = 2\cos\left( \frac{x}{2} + \frac{\pi}{4} \right)\sin\left( \frac{3x}{2} - \frac{\pi}{4} \right)\)
\(\cos^{3}x - 3\cos^{2}x + \cos x + \sin{2x} = \sin\left( x - \dfrac{\pi}{2} \right) + \sin{2x},\)
\(\cos^{3}x - 3\cos^{2}x + 2\cos x = 0,\)
\(\cos x\left( \cos^{2}x - 3\cos x + 2 \right) = 0,\)
\(\cos x = 0\) или \(\cos x = \frac{3 \pm \sqrt{9 - 8}}{2}\)
\(x_{1} = \dfrac{\pi}{2} + \pi n, n \in Z;\)
\(\cos x = 2\) – корней нет;
\(\cos x = 1,\)
\(x_{2} = 2\pi n, n \in Z;\)

3) \(\sin^{2}x + \cos^{2}{3x} = 1\)
\(\cos^{2}{3x} - \cos^{2}x = 0,\)
\(\left( \cos{3x} - \cos x \right)\left( \cos{3x} + \cos x \right) = 0,\)
\(- 2\sin{2x}\sin x = 0\) или \(2\cos{2x}\cos x = 0\)
\(\sin{2x} = 0\) или \(\sin x = 0\) или \(\cos{2x} = 0\) или \(\cos x = 0\)
\(x_{1} = \dfrac{\pi n}{2},\)
\(x_{2} = \pi n,\)
\(x_{3} = \dfrac{\pi}{4} + \dfrac{\pi n}{2},\)
\(x_{4} = \dfrac{\pi}{2} + \pi n, n \in Z;\)

Ответ: \(x_{1} = \frac{\pi n}{2}\), \(x_{2} = \frac{\pi}{4} + \frac{\pi n}{2}\), \(n \in Z\);

4) \(\operatorname{ctg} x + \sin{2x} = \operatorname{ctg} {3x}\)
\(- \dfrac{\sin{2x}}{\sin{3x}\sin x} = \sin{2x};\)

Область допустимых значений: \(\sin{3x}\sin x \neq 0\);

\(- \sin{2x}\left( 1 + \sin{3x}\sin x \right) = 0,\)
\(\sin{2x} = 0\) или \(\sin{3x}\sin x = - 1\)
\(x_{1} = \dfrac{\pi n}{2}, n \in Z;\)
\(x = \pi n\) – не входит в ОДЗ, т.к. \(\sin x = 0\); остаётся \(x_{1} = \frac{\pi}{2} + \pi n\), \(n \in Z\);
\(\cos{2x} - \cos{4x} = - 2,\)
\(\cos^{2}{2x} - \dfrac{1}{2}\cos{2x} = \dfrac{3}{2},\)
\(\left( \cos{2x} - \dfrac{1}{4} \right)^{2} = \dfrac{25}{16},\)
\(\cos{2x} = \frac{3}{2}\) – корней нет;
\(\cos{2x} = - 1,\)
\(x_{2} = - \dfrac{\pi}{2} + \pi n, n \in Z;\)

Ответ: \(x = \frac{\pi}{2} + \pi n\), \(n \in Z\).