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Lika1358
@Lika1358
July 2022
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2 cos^2 4 x+ 3 sin^2 4x+ 2 cos 4x=0 пожалуйста помогите....
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irka1804
Sin^2(4x) = 1 - cos^2(4x)
2cos^2(4x) + 3 - 3cos^2(4x) + 2cos(4x) = 0
cos^2(4x) - 2cos(4x) - 3 = 0
t = cos4x
t^2 - 2t - 3 = 0
t = -1 t = 3
t = 3 не подходит, так как cos4x <= 1
t = -1
cos4x = -1
4x = π + 2πn, n ∈ Z
x = π / 4 +
πn / 2, n
∈ Z
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Answers & Comments
2cos^2(4x) + 3 - 3cos^2(4x) + 2cos(4x) = 0
cos^2(4x) - 2cos(4x) - 3 = 0
t = cos4x
t^2 - 2t - 3 = 0
t = -1 t = 3
t = 3 не подходит, так как cos4x <= 1
t = -1
cos4x = -1
4x = π + 2πn, n ∈ Z
x = π / 4 + πn / 2, n ∈ Z