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Melada
@Melada
July 2022
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Решите, пожалуйста, подробно
7cos^2x-5sinx-5=0
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tifdoa
Переплавим cos²x на sin²x
7*(1-sin²x)-5sinx-5=0
7-7sin²x-5sinx-5=0
-7sin²x-5sinx+2=0
sinx=t
-7t²-5t+2=0
D=25-4*(-7)*2=81
t₁= (5+9) / (-14) = -1
t₂= (5-9) / (-14) = 2/7
sinx=-1
x=
n ∈ Z
sinx=2/7
x=
n ∈ Z
ответ x=
n ∈ Z
x=
n ∈ Z
3 votes
Thanks 5
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Answers & Comments
7*(1-sin²x)-5sinx-5=0
7-7sin²x-5sinx-5=0
-7sin²x-5sinx+2=0
sinx=t
-7t²-5t+2=0
D=25-4*(-7)*2=81
t₁= (5+9) / (-14) = -1
t₂= (5-9) / (-14) = 2/7
sinx=-1
x= n ∈ Z
sinx=2/7
x= n ∈ Z
ответ x= n ∈ Z
x= n ∈ Z