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ansmith15
@ansmith15
July 2022
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Решить уравнение 5/tg^2x - 19/sinx + 17=0 и определить корни, принадлежащие отрезку [-7π/2; -2π]
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sedinalana
Verified answer
5сos²x/sin²x-19/sinx+17=0
sinx≠0
5cos²x-19sinx+17sin²x=0
5-5sin²x-19sinx+17sin²x=0
sinx=a
12a²-19a+5=0
D=361-240=121
a1=(19-11)/24=1/3⇒sinx=1/3⇒x=(-1)^n*arcsin1/3+2πn.n∈z
x=-3π-arcsin1/3∈[-7π/2;-3π]
a2=(19+11)/24=1,25⇒sinx=1,25>4 нет решения
9 votes
Thanks 15
ansmith15
Да, до этого я решила, я просто не понимаю, какие корни принадлежат
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Answers & Comments
Verified answer
5сos²x/sin²x-19/sinx+17=0sinx≠0
5cos²x-19sinx+17sin²x=0
5-5sin²x-19sinx+17sin²x=0
sinx=a
12a²-19a+5=0
D=361-240=121
a1=(19-11)/24=1/3⇒sinx=1/3⇒x=(-1)^n*arcsin1/3+2πn.n∈z
x=-3π-arcsin1/3∈[-7π/2;-3π]
a2=(19+11)/24=1,25⇒sinx=1,25>4 нет решения