Вариант Б1 1 Решите уравнения: a) sin(x-π/3)+1=0; б) 1-2 cos² 2x = √2/2; в) √3 tg(π/6-x)= -1.

Вариант Б1

1. Решите уравнения:

a) $$sin(x-\frac{\pi}{3})+1=0$$

$$sin(x-\frac{\pi}{3})=-1$$ $$x-\frac{\pi}{3}=-\frac{\pi}{2}+2\pi k, k \in Z$$ $$x=-\frac{\pi}{2}+\frac{\pi}{3}+2\pi k, k \in Z$$ $$x=-\frac{3\pi}{6}+\frac{2\pi}{6}+2\pi k, k \in Z$$ $$x=-\frac{\pi}{6}+2\pi k, k \in Z$$

Ответ: $$x=-\frac{\pi}{6}+2\pi k, k \in Z$$

б) $$1-2cos^2(2x)=\frac{\sqrt{2}}{2}$$

$$-cos(4x)=\frac{\sqrt{2}}{2}-1$$ $$cos(4x)=1-\frac{\sqrt{2}}{2}$$ $$4x = \pm arccos(1-\frac{\sqrt{2}}{2})+2\pi k, k \in Z$$ $$x = \pm \frac{1}{4}arccos(1-\frac{\sqrt{2}}{2})+\frac{\pi k}{2}, k \in Z$$

Ответ: $$x = \pm \frac{1}{4}arccos(1-\frac{\sqrt{2}}{2})+\frac{\pi k}{2}, k \in Z$$

в) $$\sqrt{3}tg(\frac{\pi}{6}-x)=-1$$

$$tg(\frac{\pi}{6}-x)=-\frac{1}{\sqrt{3}}$$ $$tg(\frac{\pi}{6}-x)=-\frac{\sqrt{3}}{3}$$ $$\frac{\pi}{6}-x = arctg(-\frac{\sqrt{3}}{3})+\pi k, k \in Z$$ $$\frac{\pi}{6}-x = -\frac{\pi}{6}+\pi k, k \in Z$$ $$-x = -\frac{\pi}{6}-\frac{\pi}{6}+\pi k, k \in Z$$ $$-x = -\frac{\pi}{3}+\pi k, k \in Z$$ $$x = \frac{\pi}{3}+\pi k, k \in Z$$

Ответ: $$x = \frac{\pi}{3}+\pi k, k \in Z$$