Jawaban:
A
Penjelasan dengan langkah-langkah:
tan 4x = tan ⅓ π untuk 0 ≤ x ≤ 2π
jawab :
tan 4x = tan ⅓π, maka
4x = ⅓π + k . π
x = 1/12 π + k . ¼π
k = 0 → x = 1/12 π + 0 . ¼π = 1/12π
k = 1 → x = 1/12 π + 1 . ¼π = ⅓π
k = 2 → x = 1/12 π + 2 . ¼π = 7/12π
k = 3 → x = 1/12 π + 3 . ¼π = 5/6π
k = 4 → x = 1/12 π + 4 . ¼π = 13/12π
k = 5 → x = 1/12 π + 5 . ¼π = 4/3π
k = 6 → x = 1/12 π + 6 . ¼π = 19/12π
k = 7 → x = 1/12 π + 7 . ¼π = 11/6π
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Jawaban:
A
Penjelasan dengan langkah-langkah:
tan 4x = tan ⅓ π untuk 0 ≤ x ≤ 2π
jawab :
tan 4x = tan ⅓π, maka
4x = ⅓π + k . π
x = 1/12 π + k . ¼π
k = 0 → x = 1/12 π + 0 . ¼π = 1/12π
k = 1 → x = 1/12 π + 1 . ¼π = ⅓π
k = 2 → x = 1/12 π + 2 . ¼π = 7/12π
k = 3 → x = 1/12 π + 3 . ¼π = 5/6π
k = 4 → x = 1/12 π + 4 . ¼π = 13/12π
k = 5 → x = 1/12 π + 5 . ¼π = 4/3π
k = 6 → x = 1/12 π + 6 . ¼π = 19/12π
k = 7 → x = 1/12 π + 7 . ¼π = 11/6π