Question

я решила, но у меня не получается сделать отбор окружностью:​
помогите.

Answer 1

¯\(◉‿◉)/¯ْْْْْْْْْْْْ

Answer 2

Ответ:

$2cos\Big(\dfrac{\pi}{2}-x\Big)-tgx=0\ \ ,\ \ \ \ ODZ:\ x\ne \dfrac{\pi}{2}+\pi n\ ,\ n\in Z\\\\2sinx-\dfrac{sinx}{cosx}=0\ \ ,\ \ \ \ \dfrac{2\, sinx\cdot cosx-sinx}{cosx}=0\ \ ,\\\\sinx\cdot (2cosx-1)=0\\\\a)\ \ sinx=0\ \ ,\ \ \underline{x=\pi m\ ,\ m\in Z}\\\\b)\ \ 2cosx-1=0\ \ ,\ \ \ cosx=\dfrac{1}{2}\ \ ,\ \ \ \underline {x=\pm \dfrac{\pi}{3}+2\pi k\ ,\ k\in Z}\\\\c)\ \ x\in \Big[-2\pi \ ;-\dfrac{\pi}{2}\ \Big]:\ \ \ \underline{x_1=-2\pi \ ,\ x_2=-2\pi +\dfrac{\pi}{3}=-\dfrac{5\pi}{3}\ ,\ x_3=-\pi \ .}$

$ili:\ \ -2\pi \leq \pi m\leq -\dfrac{\pi}{2}\ \ ,\ \ \ -2\leq m\leq -\dfrac{1}{2}\ \ ,\\\\m=-2\ ,\ -1\in Z\ \ \ \ \Rightarrow \\\\m=-2:\ x=-2\pi \ \ ,\\\\m=-1:\ \ x=-\pi \\\\-2\pi \leq \pm \dfrac{\pi}{3}+2\pi k\leq -\dfrac{\pi}{2}\ \ ,\ \ \ -2\leq \pm \dfrac{1}{3}+2k\leq -\dfrac{1}{2}\ \ ,\\\\\star \ -2\leq -\dfrac{1}{3}+2k\leq -\dfrac{1}{2}\ \ ,\ \ -\dfrac{5}{3}\leq 2k\leq -\dfrac{1}{6}\ \ ,\ \ \ -\dfrac{5}{6}\leq k\leq -\dfrac{1}{12}\ ,\ k\notin Z$

$\star \ -2\leq \dfrac{1}{3}+2k\leq -\dfrac{1}{2}\ \ ,\ \ -\dfrac{7}{3}\leq 2k\leq -\dfrac{5}{6}\ \ ,\ \ \ -\dfrac{7}{6}\leq k\leq -\dfrac{5}{12}\ \ ,\\\\-1\dfrac{1}{6}\leq k\leq -\dfrac{5}{12}\ \ \ \Rightarrow \ \ \ k=-1\in Z\\\\k=-1:\ \ x=\dfrac{\pi}{3}-2\pi =-\dfrac{5\pi}{3}\ .$