Решение тригонометрических уравнений

Решим тригонометрические уравнения, представленные на доске. Для удобства разделим решение по вариантам. I вариант 1. \(3\sin^2 x - 5\sin x - 2 = 0\) Пусть \(\sin x = t\), где \(|t| \le 1\). \[3t^2 - 5t - 2 = 0\] \[D = (-5)^2 - 4 \cdot 3 \cdot (-2) = 25 + 24 = 49 = 7^2\] \[t_1 = \frac{5 + 7}{6} = 2 \text{ (не подходит, так как } 2 > 1)\] \[t_2 = \frac{5 - 7}{6} = -\frac{2}{6} = -\frac{1}{3}\] Обратная замена: \[\sin x = -\frac{1}{3}\] \[x = (-1)^{n+1} \arcsin\left(\frac{1}{3}\right) + \pi n, n \in \mathbb{Z}\] 2. \(\text{tg } 4x = 1\) \[4x = \text{arctg } 1 + \pi n\] \[4x = \frac{\pi}{4} + \pi n\] \[x = \frac{\pi}{16} + \frac{\pi n}{4}, n \in \mathbb{Z}\] 3. \(7\cos^2 x - 10\sin x \cos x + 3\sin^2 x = 0\) Это однородное уравнение второй степени. Разделим обе части на \(\cos^2 x\) (при условии \(\cos x \neq 0\)): \[3\text{tg}^2 x - 10\text{tg } x + 7 = 0\] Пусть \(\text{tg } x = t\): \[3t^2 - 10t + 7 = 0\] \[D = 100 - 4 \cdot 3 \cdot 7 = 100 - 84 = 16 = 4^2\] \[t_1 = \frac{10 + 4}{6} = \frac{14}{6} = \frac{7}{3}; \quad t_2 = \frac{10 - 4}{6} = 1\] \[\text{tg } x = \frac{7}{3} \Rightarrow x = \text{arctg}\left(\frac{7}{3}\right) + \pi n, n \in \mathbb{Z}\] \[\text{tg } x = 1 \Rightarrow x = \frac{\pi}{4} + \pi k, k \in \mathbb{Z}\] 4. \(2\cos x - 1 = 0\) \[\cos x = \frac{1}{2}\] \[x = \pm \frac{\pi}{3} + 2\pi n, n \in \mathbb{Z}\] 5. \(5\cos^2 x + 6\sin x - 6 = 0\) Используем основное тождество \(\cos^2 x = 1 - \sin^2 x\): \[5(1 - \sin^2 x) + 6\sin x - 6 = 0\] \[5 - 5\sin^2 x + 6\sin x - 6 = 0\] \[-5\sin^2 x + 6\sin x - 1 = 0\] \[5\sin^2 x - 6\sin x + 1 = 0\] Пусть \(\sin x = t\): \[5t^2 - 6t + 1 = 0\] \[D = 36 - 20 = 16 = 4^2\] \[t_1 = 1; \quad t_2 = \frac{1}{5}\] \[\sin x = 1 \Rightarrow x = \frac{\pi}{2} + 2\pi n, n \in \mathbb{Z}\] \[\sin x = \frac{1}{5} \Rightarrow x = (-1)^k \arcsin\left(\frac{1}{5}\right) + \pi k, k \in \mathbb{Z}\] II вариант 1. \(2\sin^2 x + 5\sin x + 3 = 0\) Пусть \(\sin x = t\): \[2t^2 + 5t + 3 = 0\] \[D = 25 - 24 = 1\] \[t_1 = \frac{-5 + 1}{4} = -1; \quad t_2 = \frac{-5 - 1}{4} = -1.5 \text{ (не подходит)}\] \[\sin x = -1 \Rightarrow x = -\frac{\pi}{2} + 2\pi n, n \in \mathbb{Z}\] 2. \(\sin 3x - 1 = 0\) \[\sin 3x = 1\] \[3x = \frac{\pi}{2} + 2\pi n\] \[x = \frac{\pi}{6} + \frac{2\pi n}{3}, n \in \mathbb{Z}\] 3. \(\text{tg } \frac{x}{3} = \frac{1}{\sqrt{3}}\) \[\frac{x}{3} = \text{arctg}\left(\frac{1}{\sqrt{3}}\right) + \pi n\] \[\frac{x}{3} = \frac{\pi}{6} + \pi n\] \[x = \frac{\pi}{2} + 3\pi n, n \in \mathbb{Z}\] Уравнение с первой фотографии: \(3\sin^2 x - \sin x \cos x = 2\) Заменим \(2\) на \(2(\sin^2 x + \cos^2 x)\): \[3\sin^2 x - \sin x \cos x = 2\sin^2 x + 2\cos^2 x\] \[\sin^2 x - \sin x \cos x - 2\cos^2 x = 0\] Разделим на \(\cos^2 x\): \[\text{tg}^2 x - \text{tg } x - 2 = 0\] Пусть \(\text{tg } x = t\): \[t^2 - t - 2 = 0\] По теореме Виета: \(t_1 = 2, t_2 = -1\). \[\text{tg } x = 2 \Rightarrow x = \text{arctg } 2 + \pi n, n \in \mathbb{Z}\] \[\text{tg } x = -1 \Rightarrow x = -\frac{\pi}{4} + \pi k, k \in \mathbb{Z}\]