Question

Хелпаните пжж!!!!!! ​

Answer 1

2sin2x=1-2cos²x

1-2cos²x=sin²x+cos²x-2cos²x=sin²x-cos²2x=-cos2x

2sin2x=-cos2x

2sin2x+cos2x=0

sin2x(2+ctg2x)=0   (sin2x≠0)

2+ctg2x=0

ctg2x=-2

2x=arcctg(-2)+pin (n∈Z)

x=0.5arcctg(-2)+pin/2

Answer 2

Ответ:

$4\, sinx\cdot cosx+2\, cos^2x=1\\\\4\, sinx\cdot cosx+2\, cos^2x=sin^2x+cos^2x\\\\sin^2x-4sinx\cdot cosx-cos^2x=0\ \Big|:cos^2x\ne 0\\\\tg^2x-4tgx-1=0\ \ ,\ \ \ D/4=5\ \ ,\\\\a)\ \ tgx=2-\sqrt5\ \ ,\ \ x=arctg(2-\sqrt5)+\pi n\ ,\ n\in Z\\\\b)\ \ tgx=2+\sqrt5\ \ ,\ \ x=arctg(2+\sqrt5)+\pi k\ ,\ k\in Z\\\\Otvet:\ \ x=arctg(2-\sqrt5)+\pi n\ ,\ \ x=arctg(2+\sqrt5)+\pi k\ ,\ n,k\in Z\ .$