2ctg(x)+1=tg(x-п/4)
tg(x-п/4)-2ctg(x)=1
sin(x-п/4)/cos(x-п/4)-2•cos(x)/sin(x)=1
-----
sin(x-п/4)=
=sin(x)cos(п/4)-cos(x)sin(п/4)=
=sin(x)•√2/2-cos(x)•√2/2=
=√2/2(sin(x)-cos(x))
Аналогично:
cos(x-п/4)=
=√2/2(sin(x)+cos(x))
Возвращаемся к уравнению:
(sin(x)-cos(x))/(sin(x)+cos(x))-2•cos(x)/sin(x)=1
Приводим к общему знаменателю:
(sin²x-3sin(x)cos(x)-2cos²x)/(sin²x+sin(x)cos(x))=1
sin²x-(3/2)•sin(2x)-2cos²x=sin²x+sin(x)cos(x)
-3sin(2x)-4cos²x=2sin(x)cos(x)
-3sin(2x)-4cos²x-sin(2x)=0
-4sin(2x)-4cos²x=0
-8sin(x)cos(x)-4cos²x=0
-4cos(x)(2sin(x)+cos(x))=0
Отсюда
cos(x)=0 (1)
и 2sin(x)=cos(x) (2)
(1)
cos(x)=0
x=п/2+пk
(2)
и 2sin(x)=cos(x) |:cos(x)
2tg(x)=-1 <=> tg(x)=-1/2
x=-arctg(1/2)+пk
Ответ:
x=п/2+пk, k∈Z
x=-arctg(1/2)+пk, k∈Z
Copyright © 2024 SCHOLAR.TIPS - All rights reserved.
Answers & Comments
2ctg(x)+1=tg(x-п/4)
tg(x-п/4)-2ctg(x)=1
sin(x-п/4)/cos(x-п/4)-2•cos(x)/sin(x)=1
-----
sin(x-п/4)=
=sin(x)cos(п/4)-cos(x)sin(п/4)=
=sin(x)•√2/2-cos(x)•√2/2=
=√2/2(sin(x)-cos(x))
-----
Аналогично:
cos(x-п/4)=
=√2/2(sin(x)+cos(x))
-----
Возвращаемся к уравнению:
(sin(x)-cos(x))/(sin(x)+cos(x))-2•cos(x)/sin(x)=1
Приводим к общему знаменателю:
(sin²x-3sin(x)cos(x)-2cos²x)/(sin²x+sin(x)cos(x))=1
sin²x-(3/2)•sin(2x)-2cos²x=sin²x+sin(x)cos(x)
-3sin(2x)-4cos²x=2sin(x)cos(x)
-3sin(2x)-4cos²x-sin(2x)=0
-4sin(2x)-4cos²x=0
-8sin(x)cos(x)-4cos²x=0
-4cos(x)(2sin(x)+cos(x))=0
Отсюда
cos(x)=0 (1)
и 2sin(x)=cos(x) (2)
(1)
cos(x)=0
x=п/2+пk
(2)
и 2sin(x)=cos(x) |:cos(x)
2tg(x)=-1 <=> tg(x)=-1/2
x=-arctg(1/2)+пk
Ответ:
x=п/2+пk, k∈Z
x=-arctg(1/2)+пk, k∈Z