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stoneddeer6
@stoneddeer6
July 2022
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cos^2(5x)+7sin^2(5x)=8 sin5x cos5x
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m11m
cos² 5x
+
7sin² 5x
-
8sin5x cos5x
=
0
cos² 5x cos² 5x cos² 5x cos² 5x
1 + 7tg² 5x - 8tg 5x =0
Замена tg 5x =y
7y² -8y +1=0
D=64 -28 =36
y₁ =
8-6
= 2/14 = 1/7
14
y₂ =
8+6
= 1
14
При у= 1/7
tg 5x=1/7
5x= arctg 1/7 + πk, k∈Z
x= 0.2 arctg1/7 + 0.2πk , k∈Z
При у=1
tg 5x=1
5x= π/4 + πk, k∈Z
x= π/20 + 0.2πk, k∈Z
Ответ: 0,2arctg 1/7 + 0.2πk, k∈Z;
π/20 + 0.2πk, k∈Z
0 votes
Thanks 4
stoneddeer6
тоесть мы поделили на cos^2(5x) после этого 5х все равно остаётся?
m11m
да
stoneddeer6
хорошо,спасибо
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Answers & Comments
cos² 5x cos² 5x cos² 5x cos² 5x
1 + 7tg² 5x - 8tg 5x =0
Замена tg 5x =y
7y² -8y +1=0
D=64 -28 =36
y₁ = 8-6 = 2/14 = 1/7
14
y₂ = 8+6 = 1
14
При у= 1/7
tg 5x=1/7
5x= arctg 1/7 + πk, k∈Z
x= 0.2 arctg1/7 + 0.2πk , k∈Z
При у=1
tg 5x=1
5x= π/4 + πk, k∈Z
x= π/20 + 0.2πk, k∈Z
Ответ: 0,2arctg 1/7 + 0.2πk, k∈Z;
π/20 + 0.2πk, k∈Z