Question

Помогите пожалуйста с тригонометрическими уравнениями

Answer 1

$1) \ \text{tg} \, 3x (2\cos x - \sqrt{2}) = 0$

$\left[\begin{array}{ccc}\text{tg} \, 3x = 0, \ \ \ \ \ \ \ \ \ \ \ \ (1) \\2\cos x - \sqrt{2} = 0 \ \ \ \ (2) \\\end{array}\right$

$(1)\ \text{tg} \, 3x = 0$

$3x = \text{arctg} \, 0 + \pi n, \ n \in Z$

$3x = \pi n, \ n \in Z$

$x = \dfrac{\pi n}{3}, \ n \in Z$

$(2) \ 2\cos x - \sqrt{2} = 0$

$\cos x = \dfrac{\sqrt{2}}{2}$

$x = \pm \arccos \dfrac{\sqrt{2}}{2} + 2\pi k, \ k \in Z$

$x = \pm \dfrac{\pi}{4} + 2\pi k, \ k \in Z$

Ответ: $x = \dfrac{\pi n}{3}, \ n \in Z; \ x = \pm \dfrac{\pi}{4} + 2\pi k, \ k \in Z$

$2) \ \cos^{2} 2x + \sin^{2}2x + \sin x - 1 = -\dfrac{\sqrt{3}}{2}$

$1 + \sin x - 1 = -\dfrac{\sqrt{3}}{2}$

$\sin x = -\dfrac{\sqrt{3}}{2}$

$x = (-1)^{n}\arcsin \left(-\dfrac{\sqrt{3}}{2} \right) + \pi n, \ n \in Z$

$x = (-1)^{n} \cdot (-1) \cdot \arcsin \dfrac{\sqrt{3}}{2} + \pi n, \ n \in Z$

$x = (-1)^{n+1}\dfrac{\pi}{3} + \pi n, \ n \in Z$

Ответ: $x = (-1)^{n+1}\dfrac{\pi}{3} + \pi n, \ n \in Z$

$3) \ 1 + \cos x = 2 - \sin^{2}x$

$1 + \cos x - 2 + \sin^{2}x = 0$

$\cos x - 1 + 1 - \cos^{2}x = 0$

$\cos x - \cos^{2}x = 0$

$\cos x(1 - \cos x) = 0$

$\left[\begin{array}{ccc}\cos x = 0, \ \ \ \ \ \ \ \ (1) \\1 - \cos x = 0 \ \ \ \ (2) \\\end{array}\right$

$(1) \ \cos x = 0$

$x = \dfrac{\pi}{2} + \pi n, \ n \in Z$

$(2) \ 1 - \cos x = 0$

$\cos x =1$

$x = 2\pi k, \ k \in Z$

Ответ: $x = \dfrac{\pi}{2} + \pi n, \ n \in Z; \ x = 2\pi k, \ k \in Z$