Question

Уравнения. 90 баллов.

Answer 1

Ответ:

$4)\ \ 4tg^2x-5tgx=0\\\\tgx(4tgx-5)=0\\\\a)\ \ tgx=0\ \ ,\ \ x=\pi n\ ,\ n\in Z\\\\b)\ \ tgx=1,25\ \ ,\ \ x=arctg1,25+\pi k\ ,\ k\in Z\\\\Otvet:\ \ x_1=\pi n\ \ ,\ x_2=arctg1,25+\pi k\ \ ,\ n,k\in Z\ .$

$5)\ \ 8tgx-9ctgx+1=0\\\\t=tgx\ ,\ \ ctgx=\dfrac{1}{t}\ \ ,\ \ \ 8t-\dfrac{9}{t}+1=0\ \ ,\ \ \dfrac{8t^2+t-9}{t}=0\ \ ,\\\\8t^2+t-9=0\ \ ,\ \ D=289=17^2\ \ ,\ \ t_1=-\dfrac{9}{2}\ ,\ t_2=1\ ,\\\\a)\ \ tgx=-4,5\ \ ,\ \ x=-arctg4,5+\pi n\ ,\ n\in Z\\\\b)\ \ tgx=1\ \ ,\ \ x=\dfrac{\pi }{4}+\pi k\ ,\ k\in Z\\\\Otvet:\ \ x_1=-arctg4,5+\pi n\ \ ,\ \ x_2=\dfrac{\pi }{4}+\pi k\ ,\ n,k\in Z\ .$

$\displaystyle 6)\ \ 4sin\dfrac{x}{5}=5\, cos\frac{x}{5}\ \ \Big|:cos\frac{x}{5}\ne 0\\\\4\, tg\frac{x}{5}=5\ \ ,\ \ \ tg\frac{x}{5}=1,25\ \ ,\ \ \ \frac{x}{5}=arctg1,25+\pi n\ \ ,\\\\x=5\, arctg1,25+5\pi n\ \ ,\ n\in Z\\\\Otvet:\ \ x=5\, arctg1,25+5\pi n\ \ ,\ n\in Z\ .$