Предмет: Алгебра, автор: kambul22

Решите
На отрезке {3П/2;5П/2]
7sin ^{2} x-4sinxcosx-3cos ^{2} x=0

Ответы

Автор ответа: nafanya2014
0
Это однородное тригонометрическое уравнение, решается делением обеих частей на
cos²x≠0  ( или sin²x≠0).
Получим квадратное уравнение относительно tgx
7tg²x-4tgx-3=0
D=(-4)²-4·7·(-3)=16+84=100=10²
tgx=(4+10)/14=1          или       tgx=(4-10)/14=-6/14=-3/7
x=arctg1+πk, k∈Z        или       x=arctg(-3/7)+πn, n∈Z
x=(π/4)+πk, k∈Z
Указанному промежутку [3π/2; 5π/2] принадлежат
(π/4)+2π=9π/4
и
-arctg (3/7) +2π

Ответ. (π/4)+2π=9π/4;  -arctg (3/7) +2π






kazimierz2015: супер
kambul22: ответ не такой там
nafanya2014: напишите, какой там ответ
kambul22: (-1)^n П/4+Пn П/2+Пn
nafanya2014: Это неверный ответ. При х=П/2 получаем 7-0-0=0 - неверно
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