Вариант 1 1. Решить тригонометрические уравнения: a) 2cosx+√2=0 б) tg2x+1=0 B) sin(x+4)=-1 r) 2cos2x-3cosx+1=0 д) sin3x+√3cos3x=0 e) 2tgx-ctgx+1=0 ж) 3cos2x+sinxcosx-2sin2x=0

Решение тригонометрических уравнений:

1. а) $$2\cos x + \sqrt{2} = 0$$ $$\cos x = -\frac{\sqrt{2}}{2}$$ $$x = \pm \frac{3\pi}{4} + 2\pi n, n \in \mathbb{Z}$$ Ответ: $$x = \pm \frac{3\pi}{4} + 2\pi n, n \in \mathbb{Z}$$
2. б) $$\text{tg } 2x + 1 = 0$$ $$\text{tg } 2x = -1$$ $$2x = -\frac{\pi}{4} + \pi n, n \in \mathbb{Z}$$ $$x = -\frac{\pi}{8} + \frac{\pi}{2} n, n \in \mathbb{Z}$$ Ответ: $$x = -\frac{\pi}{8} + \frac{\pi}{2} n, n \in \mathbb{Z}$$
3. в) $$\sin(x+\frac{\pi}{4}) = -1$$ $$x + \frac{\pi}{4} = -\frac{\pi}{2} + 2\pi n, n \in \mathbb{Z}$$ $$x = -\frac{\pi}{2} - \frac{\pi}{4} + 2\pi n, n \in \mathbb{Z}$$ $$x = -\frac{3\pi}{4} + 2\pi n, n \in \mathbb{Z}$$ Ответ: $$x = -\frac{3\pi}{4} + 2\pi n, n \in \mathbb{Z}$$
4. г) $$2\cos 2x - 3\cos x + 1 = 0$$ $$2(2\cos^2 x - 1) - 3\cos x + 1 = 0$$ $$4\cos^2 x - 2 - 3\cos x + 1 = 0$$ $$4\cos^2 x - 3\cos x - 1 = 0$$ Пусть $$t = \cos x$$, тогда $$4t^2 - 3t - 1 = 0$$ $$D = (-3)^2 - 4 \cdot 4 \cdot (-1) = 9 + 16 = 25$$ $$t_1 = \frac{3 + 5}{8} = \frac{8}{8} = 1$$ $$t_2 = \frac{3 - 5}{8} = \frac{-2}{8} = -\frac{1}{4}$$ $$\cos x = 1$$\Rightarrow $$x = 2\pi n, n \in \mathbb{Z}$$ $$\cos x = -\frac{1}{4}$$ $$x = \pm \arccos(-\frac{1}{4}) + 2\pi n, n \in \mathbb{Z}$$ Ответ: $$x = 2\pi n, x = \pm \arccos(-\frac{1}{4}) + 2\pi n, n \in \mathbb{Z}$$
5. д) $$\sin 3x + \sqrt{3} \cos 3x = 0$$ $$\sin 3x = -\sqrt{3} \cos 3x$$ $$\text{tg } 3x = -\sqrt{3}$$ $$3x = -\frac{\pi}{3} + \pi n, n \in \mathbb{Z}$$ $$x = -\frac{\pi}{9} + \frac{\pi}{3} n, n \in \mathbb{Z}$$ Ответ: $$x = -\frac{\pi}{9} + \frac{\pi}{3} n, n \in \mathbb{Z}$$
6. е) $$2 \text{tg } x - \text{ctg } x + 1 = 0$$ $$2 \frac{\sin x}{\cos x} - \frac{\cos x}{\sin x} + 1 = 0$$ $$\frac{2 \sin^2 x - \cos^2 x + \sin x \cos x}{\sin x \cos x} = 0$$ $$2 \sin^2 x - \cos^2 x + \sin x \cos x = 0$$ $$2 \sin^2 x - (1 - \sin^2 x) + \sin x \cos x = 0$$ $$3 \sin^2 x + \sin x \cos x - 1 = 0$$ Разделим обе части уравнения на $$\cos^2 x$$ $$3 \text{tg}^2 x + \text{tg } x - \frac{1}{\cos^2 x} = 0$$ $$3 \text{tg}^2 x + \text{tg } x - (1 + \text{tg}^2 x) = 0$$ $$2 \text{tg}^2 x + \text{tg } x - 1 = 0$$ Пусть $$t = \text{tg } x$$, тогда $$2t^2 + t - 1 = 0$$ $$D = 1^2 - 4 \cdot 2 \cdot (-1) = 1 + 8 = 9$$ $$t_1 = \frac{-1 + 3}{4} = \frac{2}{4} = \frac{1}{2}$$ $$t_2 = \frac{-1 - 3}{4} = \frac{-4}{4} = -1$$ $$\text{tg } x = \frac{1}{2}$$ $$x = \text{arctg } \frac{1}{2} + \pi n, n \in \mathbb{Z}$$ $$\text{tg } x = -1$$ $$x = -\frac{\pi}{4} + \pi n, n \in \mathbb{Z}$$ Ответ: $$x = \text{arctg } \frac{1}{2} + \pi n, x = -\frac{\pi}{4} + \pi n, n \in \mathbb{Z}$$
7. ж) $$3\cos 2x + \sin x \cos x - 2 \sin 2x = 0$$ $$3(\cos^2 x - \sin^2 x) + \sin x \cos x - 4 \sin x \cos x = 0$$ $$3\cos^2 x - 3\sin^2 x - 3 \sin x \cos x = 0$$ Разделим обе части уравнения на $$\cos^2 x$$ $$3 - 3 \text{tg}^2 x - 3 \text{tg } x = 0$$ $$\text{tg}^2 x + \text{tg } x - 1 = 0$$ Пусть $$t = \text{tg } x$$, тогда $$t^2 + t - 1 = 0$$ $$D = 1^2 - 4 \cdot 1 \cdot (-1) = 1 + 4 = 5$$ $$t_1 = \frac{-1 + \sqrt{5}}{2}$$ $$t_2 = \frac{-1 - \sqrt{5}}{2}$$ $$\text{tg } x = \frac{-1 + \sqrt{5}}{2}$$ $$x = \text{arctg } (\frac{-1 + \sqrt{5}}{2}) + \pi n, n \in \mathbb{Z}$$ $$\text{tg } x = \frac{-1 - \sqrt{5}}{2}$$ $$x = \text{arctg } (\frac{-1 - \sqrt{5}}{2}) + \pi n, n \in \mathbb{Z}$$ Ответ: $$x = \text{arctg } (\frac{-1 + \sqrt{5}}{2}) + \pi n, x = \text{arctg } (\frac{-1 - \sqrt{5}}{2}) + \pi n, n \in \mathbb{Z}$$