Question

Решите пожалуйста!!!!!!!!!!

40 БАЛЛОВ!!!

Answer 1

Ответ:

$1)\ \ a)\ \ tgx=-\sqrt3\ \ ,\ \ x=-\dfrac{\pi}{3}+\pi n\ ,\ n\in Z\\\\b)\ \ ctgx=0\ \ ,\ \ \ x=\dfrac{\pi}{2}+\pi k\ ,\ k\in Z\\\\c)\ \ (tg2x-1)(2cosx+1)=0\\\\tg2x=1\ \ ,\ \ 2x=\dfrac{\pi}{4}+\pi n\ \ ,\ \ x_1=\dfrac{\pi}{8}+\dfrac{\pi n}{2}\ ,\ n\in Z\\\\cosx=-\dfrac{1}{2}\ \ ,\ \ x_2=\dfrac{2\pi }{3}+2\pi k\ ,\ k\in Z$

$2)\ \ 3\, ctg3x-\sqrt3=0\ \ \ \to \ \ \ ctg3x=\dfrac{\sqrt3}{3}\ \ ,\ \ 3x=\dfrac{\pi}{3}+\pi n\ ,\ n\in Z\\\\\\x=\dfrac{\pi}{9}+\dfrac{\pi n}{3}\ \ ,\ \ n\in Z\\\\\\x\in \Big[\ \dfrac{\pi}{6}\ ;\ \pi \ \Big]\ :\ x_1=\dfrac{\pi}{9}+\dfrac{\pi}{3}=\dfrac{4\pi}{9}\ \ ,\ \ x_2=\dfrac{\pi}{9}+\dfrac{2\pi}{3}=\dfrac{7\pi }{9}$

$3)\ \ 3\, ctg2x\geq 2\ \ \ \to \ \ \ ctg2x\geq \dfrac{2}{3}\ \ ,\\\\\\\pi n<2x\leq arcctg\dfrac{2}{3}+\pi n\ \ ,\ n\in Z\\\\\\\dfrac{\pi n}{2}<x\leq \dfrac{1}{2}\, arctg\dfrac{2}{3}+\dfrac{\pi n}{2}\ \ ,\ n\in Z\\\\\\x\in \Big(\ \dfrac{\pi n}{2}\ ;\ \dfrac{1}{2}\, arctg\dfrac{2}{3}+\dfrac{\pi n}{2}\ \Big]\ \ ,\ n\in Z$