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Katrinshark99
@Katrinshark99
August 2022
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Решите пожалуйста:
1) √3 cos3x = sin3x
2) 2 sin( π - 3x) + cos( 2π - 3x) = 0
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Dимасuk
Verified answer
√3cos3x = sin3x (cos3x ≠ 0)
√3 = tg3x
3x = π/3 + πn, n ∈ Z
x = π/9 + πn/3, n ∈ Z
2sin(π - 3x) + cos(2π - 3x) = 0
-2sin(-3x) + cos(3x - 2π) = 0
2sin3x + cos3x = 0
2sin3x = -cos3x
2tg3x = -1
tg3x = -1/2
3x = arctg(-1/2) + πn, n ∈ Z
x = (arctg(-1/2))/3 + πn/3, n ∈ Z.
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Answers & Comments
Verified answer
√3cos3x = sin3x (cos3x ≠ 0)√3 = tg3x
3x = π/3 + πn, n ∈ Z
x = π/9 + πn/3, n ∈ Z
2sin(π - 3x) + cos(2π - 3x) = 0
-2sin(-3x) + cos(3x - 2π) = 0
2sin3x + cos3x = 0
2sin3x = -cos3x
2tg3x = -1
tg3x = -1/2
3x = arctg(-1/2) + πn, n ∈ Z
x = (arctg(-1/2))/3 + πn/3, n ∈ Z.