Страница 412 номер 1385, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(\operatorname{tg} {2x} = 3\operatorname{tg} x\)
\(\dfrac{\sin{2x}}{\cos{2x}} = \dfrac{3\sin x}{\cos x},\)
\(\dfrac{2\sin x\cos^{2}x - 3\sin x\left( \cos^{2}x - \sin^{2}x \right)}{\cos{2x}\cos x} = 0;\)

Область допустимых значений: \(\cos{2x}\cos x \neq 0\), \(x \neq \frac{\pi}{4} + \frac{\pi}{2}n\),

\(x \neq \dfrac{\pi}{2} + \pi n, n \in Z;\)
\(3\sin^{3}x - \sin x\cos^{2}x = 0,\)
\(\sin x\left( 3 - 3\cos^{2}x - \cos^{2}x \right) = 0,\)
\(\sin x = 0\) или \(4\cos^{2}x = 3\)
\(x_{1} = \pi n, n \in Z;\)
\(\cos x = \pm \dfrac{\sqrt{3}}{2};\)
\(x_{2} = \pm \dfrac{\pi}{6} + 2\pi n,\)
\(x_{3} = \pm \dfrac{5\pi}{6} + 2\pi n, n \in Z;\)

2) \(\operatorname{ctg} {2x} = 2\operatorname{ctg} x\)
\(\dfrac{\cos{2x}}{\sin{2x}} = \dfrac{2\cos x}{\sin x},\)
\(\dfrac{\cos{2x}}{2\sin x\cos x} - \dfrac{4\cos^{2}x}{2\sin x\cos x} = 0,\)

Область допустимых значений: \(\sin{2x} \neq 0\), \(x \neq \frac{\pi}{2}n\), \(n \in Z\);

\(\cos^{2}x - \sin^{2}x - 4\cos^{2}x = 0,\)
\(3\cos^{2}x + 1 - \cos^{2}x = 0,\)
\(2\cos^{2}x = - 1,\)
\(\cos^{2}x = - \frac{1}{2}\) – корней нет;

3) \(\operatorname{tg} \left( x + \frac{\pi}{4} \right) + \operatorname{tg} \left( x - \frac{\pi}{4} \right) = 2\)
\(\dfrac{\operatorname{tg} x + \operatorname{tg} \frac{\pi}{4}}{1 - \operatorname{tg} x\operatorname{tg} \dfrac{\pi}{4}} + \dfrac{\operatorname{tg} x - \operatorname{tg} \frac{\pi}{4}}{1 + \operatorname{tg} x\operatorname{tg} \dfrac{\pi}{4}} = 2,\)
\(\dfrac{\operatorname{tg} x + 1}{1 - \operatorname{tg} x} + \dfrac{\operatorname{tg} x - 1}{1 + \operatorname{tg} x} = 2,\)
\(\dfrac{\left( 1 + \operatorname{tg} x \right)^{2} - \left( 1 - \operatorname{tg} x \right)^{2}}{1 - \operatorname{tg}^{2}x} = 2,\)
\(\dfrac{4\operatorname{tg} x}{1 - \operatorname{tg}^{2}x} = 2,\)
\(2\operatorname{tg} (2x) = 2,\)
\(2x = \dfrac{\pi}{4} + \pi n,\)
\(\cos\left( x + \frac{\pi}{4} \right)\cos\left( x - \frac{\pi}{4} \right) = \frac{1}{2}\cos{2x} \neq 0\) – левая часть исходного уравнения имеет смысл;
\(x = \dfrac{\pi}{8} + \dfrac{\pi}{2}n, n \in Z;\)

4) \(\operatorname{tg} (2x + 1)\operatorname{ctg} (x + 1) = 1\)
\(\operatorname{tg} (2x + 1) = \operatorname{tg} (x + 1),\)
\(\operatorname{tg} (2x + 1) - \operatorname{tg} (x + 1) = 0,\)
\(\dfrac{\sin(2x + 1 - x - 1)}{\cos(2x + 1)\cos(x + 1)} = 0,\)

Область допустимых значений:

\(\cos(2x + 1)\cos(x + 1) \neq 0,\ \sin(x + 1) \neq 0;\)
\(\sin x = 0,\)
\(x = \pi n, n \in Z.\)