1.hasil dari cos 2x=0 ,dengan 0≤x≤2Π 2. hsil dari √3 tan 3x +√3=0, dengan 0≤x≤2Π
Matematika
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Pertanyaan
1.hasil dari cos 2x=0 ,dengan 0≤x≤2Π
2. hsil dari √3 tan 3x +√3=0, dengan 0≤x≤2Π
2. hsil dari √3 tan 3x +√3=0, dengan 0≤x≤2Π
2 Jawaban
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1. Jawaban Anonyme
Bab Trigonometri
Matematika SMA Kelas X
1] cos 2x = 0
2x = π/2
x = π/4
2x = 3π/2
x = 3π/4
2x = 5π/2
x = 5π/4
2x = 7π/2
x = 7π/4
HP = {π/4, 3π/4, 5π/4, 7π/4}
2] √3 tan 3x + √3 = 0
√3 tan 3x = -√3
tan 3x = -√3/√3
tan 3x = - 1
3x = 3π/4
x = π/4
3x = 7π/4
x = 7π/12
3x = 11π/4
x = 11π/12
3x = 15π/4
x = 5π/4
3x = 19π/4
x = 19π/12
3x = 23π/4
x = 23π/12
HP = { π/4, 7π/12, 11π/12, 5π/4, 19π/12. 23π/12 } -
2. Jawaban Sutr1sn0
1]
cos 2x = 0
2x = phi/2
x = phi/4
2x = 3phi/2
x = 3phi/4
2x = 5phi/2
x = 5phi/4
2x = 7phi/2
x = 7phi/4
Jadi HPnya adalah {(phi/4 , 3phi/4 , 5phi/4 , 7phi/4) }
2]
V3 tan 3x + V3 = 0
V3 tan 3x = -V3
tan 3x = -V3/V3
tan 3x = -1
3x = 3phi/4
x = 3phi/12
x = phi/4
3x = 7phi/4
x = 7phi/12
3x = 11phi/4
x = 11phi/12
3x = 15phi/4
x = 15phi/12
x = 5phi/4
3x = 19phi/4
x = 19phi/12
3x = 23phi/4
x = 23phi/12
Jadi HPnya adalah {(phi/4 , 7phi/12 , 11phi/12 , 5phi/4 , 19phi/12 , 23phi/12) }