Question

1+ctg2x=1/cos(3pi/2-2x) на промежутке -2pi ; -pi/2

Answer 1

1+ctg2x=1/-sin2x
sin2x+cos2x=-1
2sinxcosx+cos²x-sin²x+sin²x+cos²x=0
2sinxcosx+2cos²x=0 /cos²x≠0
2tgx+2=0
tgx=-1
x=-π/4+πn
x=-5π/4∈[-2π;-π/2]