Вариант Б2 a) cos(x+π/6)-1=0; б) sin(x/4) cos(x/4)= -1/4; в) √3ctg(π/3-x)= -3.

Вариант Б2

a) $$cos(x+\frac{\pi}{6})-1=0$$

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

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

б) $$sin(\frac{x}{4}) cos(\frac{x}{4})= -\frac{1}{4}$$

$$\frac{1}{2}*2sin(\frac{x}{4}) cos(\frac{x}{4})= -\frac{1}{4}$$или

$$\frac{x}{2}=\pi-arcsin(-\frac{1}{2})+\pi k, k \in Z$$ $$\frac{x}{2}=\pi+\frac{\pi}{6}+\pi k, k \in Z$$ $$\frac{x}{2}=\frac{7\pi}{6}+ \pi k, k \in Z$$ $$x=\frac{7\pi}{3}+2\pi k, k \in Z$$

Ответ: $$x=-\frac{\pi}{3}+2\pi k, k \in Z; x=\frac{7\pi}{3}+2\pi k, k \in Z$$

в) $$\sqrt{3}ctg(\frac{\pi}{3}-x)= -3$$

$$ctg(\frac{\pi}{3}-x)= -\frac{3}{\sqrt{3}}$$ $$ctg(\frac{\pi}{3}-x)= -\sqrt{3}$$ $$\frac{\pi}{3}-x=arcctg(-\sqrt{3})+\pi k, k \in Z$$ $$\frac{\pi}{3}-x=\frac{5\pi}{6}+\pi k, k \in Z$$ $$-x=\frac{5\pi}{6}-\frac{\pi}{3}+\pi k, k \in Z$$ $$-x=\frac{5\pi}{6}-\frac{2\pi}{6}+\pi k, k \in Z$$ $$-x=\frac{3\pi}{6}+\pi k, k \in Z$$ $$-x=\frac{\pi}{2}+\pi k, k \in Z$$ $$x=-\frac{\pi}{2}+\pi k, k \in Z$$

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