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Решение тригонометрических уравнений (Вариант 3)
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Представлено решение тригонометрических уравнений, включающих нахождение x для ctg, tg, sin и cos функций. Приведены общие формулы и ответы для каждого уравнения.
Подробное решение
Решение тригонометрических уравнений (Вариант 3)
1) \(\text{ctg } x = -1\)
\[x = \text{arcctg}(-1) + \pi n, n \in \mathbb{Z}\]
\[x = \frac{3\pi}{4} + \pi n, n \in \mathbb{Z}\]
2) \(\text{tg } x = \frac{\sqrt{3}}{3}\)
\[x = \text{arctg}\left(\frac{\sqrt{3}}{3}\right) + \pi n, n \in \mathbb{Z}\]
\[x = \frac{\pi}{6} + \pi n, n \in \mathbb{Z}\]
3) \(\text{ctg } x = \frac{3}{2}\)
\[x = \text{arcctg}\left(\frac{3}{2}\right) + \pi n, n \in \mathbb{Z}\]
4) \(\text{tg } x = -3\)
\[x = \text{arctg}(-3) + \pi n, n \in \mathbb{Z}\]
\[x = -\text{arctg}(3) + \pi n, n \in \mathbb{Z}\]
5) \(\cos 3x = \frac{\sqrt{2}}{2}\)
\[3x = \pm \frac{\pi}{4} + 2\pi n, n \in \mathbb{Z}\]
\[x = \pm \frac{\pi}{12} + \frac{2\pi n}{3}, n \in \mathbb{Z}\]
6) \(\sin \frac{2x}{3} = \frac{\sqrt{2}}{5}\)
\[\frac{2x}{3} = (-1)^n \arcsin\left(\frac{\sqrt{2}}{5}\right) + \pi n, n \in \mathbb{Z}\]
\[x = \frac{3}{2} (-1)^n \arcsin\left(\frac{\sqrt{2}}{5}\right) + \frac{3\pi n}{2}, n \in \mathbb{Z}\]
7) \(\text{ctg } \frac{x}{4} = -\sqrt{3}\)
\[\frac{x}{4} = \text{arcctg}(-\sqrt{3}) + \pi n, n \in \mathbb{Z}\]
\[\frac{x}{4} = \frac{5\pi}{6} + \pi n, n \in \mathbb{Z}\]
\[x = \frac{10\pi}{3} + 4\pi n, n \in \mathbb{Z}\]
8) \(\cos(-2x) = -\frac{1}{3}\)
Так как \(\cos(-2x) = \cos(2x)\):
\[2x = \pm \arccos\left(-\frac{1}{3}\right) + 2\pi n, n \in \mathbb{Z}\]
\[2x = \pm (\pi - \arccos\frac{1}{3}) + 2\pi n, n \in \mathbb{Z}\]
\[x = \pm \left(\frac{\pi}{2} - \frac{1}{2}\arccos\frac{1}{3}\right) + \pi n, n \in \mathbb{Z}\]
9) \(\sin\left(\frac{\pi}{6} - x\right) = \frac{1}{2}\)
\[\frac{\pi}{6} - x = (-1)^k \frac{\pi}{6} + \pi k, k \in \mathbb{Z}\]
\[-x = -\frac{\pi}{6} + (-1)^k \frac{\pi}{6} + \pi k, k \in \mathbb{Z}\]
\[x = \frac{\pi}{6} - (-1)^k \frac{\pi}{6} - \pi k, k \in \mathbb{Z}\]
10) \(\cos\left(\frac{x}{3} + \frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2}\)
\[\frac{x}{3} + \frac{\pi}{4} = \pm \frac{3\pi}{4} + 2\pi n, n \in \mathbb{Z}\]
\[\frac{x}{3} = -\frac{\pi}{4} \pm \frac{3\pi}{4} + 2\pi n, n \in \mathbb{Z}\]
\[x = -\frac{3\pi}{4} \pm \frac{9\pi}{4} + 6\pi n, n \in \mathbb{Z}\]
11) \(\text{tg}\left(2x - \frac{\pi}{3}\right) = -\sqrt{3}\)
\[2x - \frac{\pi}{3} = -\frac{\pi}{3} + \pi n, n \in \mathbb{Z}\]
\[2x = 0 + \pi n, n \in \mathbb{Z}\]
\[x = \frac{\pi n}{2}, n \in \mathbb{Z}\]
12) \(4\text{ctg } 2x + \text{ctg}^2 2x - 5 = 0\)
Пусть \(\text{ctg } 2x = t\):
\[t^2 + 4t - 5 = 0\]
По теореме Виета: \(t_1 = 1, t_2 = -5\).
1) \(\text{ctg } 2x = 1 \Rightarrow 2x = \frac{\pi}{4} + \pi n \Rightarrow x = \frac{\pi}{8} + \frac{\pi n}{2}, n \in \mathbb{Z}\)
2) \(\text{ctg } 2x = -5 \Rightarrow 2x = \text{arcctg}(-5) + \pi k \Rightarrow x = \frac{1}{2}\text{arcctg}(-5) + \frac{\pi k}{2}, k \in \mathbb{Z}\)