Решите уравнения (х+4)(2х-1)=х(3х+11).

Question

Алгебра: Решите уравнения (х+4)(2х-1)=х(3х+11). ​

marijasoe19gmailcon · Accepted Answer

1)Для сos(a)=0,8:
sin^2(a) = 1 - cos^2(a) = 1 - 0,64 = 0,36
По условию 0<a<pi/2 => a лежит в первой четверти => sin, cos, tg и ctg - положительные
sin(a)=0,6
tg(a)=sin(a)/cos(a)=0,6/0,8=3/4
ctg(a)=4/3
2) sin(a) = 5/13
cos^2(a) = 1 - sin^2(a) = 1 - 25/169 = 144/169
cos(a) = 12/13
tg(a) = (5/13)*(13/12) = 5/12
ctg(a) = 12/5
3) tg(a) = 2,4
tg^2(a) = 5,76
1 + tg^2(a) = 1/cos^2(a) => 1/cos^2(a) = 6,76
cos^2(a) = 1/6,76 = 100/676
cos(a) = 10/26 = 5/13
sin^2(a) = 1 - 25/169 = 144/169
sin(a) = 12/13
ctg(a) = (5/13)*(13/12) = 5/12
4) ctg(a) = 7/24
tg(a) = 24/7
ctg^2(a) = 49/576
1+ctg^2(a) = 1/sin^2(a) => 1/sin^2(a) = 1 + 49/576 = 625/576
sin^2(a) = 576/625
sin(a) = 24/25
cos^2(a) = 1 - sin^2(a) = 1 - 576/625 = 49/625
cos(a) = 7/25

Решите уравнения (х+4)(2х-1)=х(3х+11). ​

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