Pertanyaan

Tentukan himpunan penyelesaian dari persamaan-persamaan trigonometri berikut! a. sin 7 x = sin 3 1 π , 0 ≤ x ≤ 2 π

Tentukan himpunan penyelesaian dari persamaan-persamaan trigonometri berikut!

a. $sin 7 x = sin \frac{1}{3} π, 0 \leq x \leq 2 π$

E. Lestari

Master Teacher

Mahasiswa/Alumni Universitas Sebelas Maret

Jawaban terverifikasi

Jawaban

himpunan penyelesaiannya adalah .

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Pembahasan

Jawaban yang benar untuk pertanyaan tersebut adalah { 21 1 π , 21 2 π , 21 7 π , 21 8 π , 21 13 π , 21 14 π , 21 19 π , 21 20 π , 21 25 π , 21 26 π , 21 31 π , 21 32 π , 21 37 π , 21 38 π } . Persamaan Trigonometri Dasar Ingat persamaan trigonometri dasar berikut: Jika sin x = sin α , nilai x = α + k ⋅ 2 π atau x = ( π − α ) + k ⋅ 2 π . sin 7 x = sin 3 1 π , 0 ≤ x ≤ 2 π sin 7 x 7 x x = = = sin 3 1 π 3 1 π + k ⋅ 2 π 21 1 π + k ⋅ 7 2 π atau sin 7 x 7 x 7 x x = = = = sin 3 1 π ( π − 3 1 π ) + k ⋅ 2 π 3 2 π + k ⋅ 2 π 21 2 π + k ⋅ 7 2 π Diperoleh: k = 0 → x = 21 1 π + 0 ⋅ 7 2 π = 21 1 π ( M ) atau k = 0 → x = 21 2 π + 0 ⋅ 7 2 π = 21 2 π ( M ) k = 1 → x = 21 1 π + 1 ⋅ 7 2 π = 21 7 π ( M ) atau k = 1 → x = 21 2 π + 1 ⋅ 7 2 π = 21 8 π ( M ) k = 2 → x = 21 1 π + 2 ⋅ 7 2 π = 21 13 π ( M ) atau k = 2 → x = 21 2 π + 2 ⋅ 7 2 π = 21 14 π ( M ) k = 3 → x = 21 1 π + 3 ⋅ 7 2 π = 21 19 π ( M ) atau k = 3 → x = 21 2 π + 3 ⋅ 7 2 π = 21 20 π ( M ) k = 4 → x = 21 1 π + 4 ⋅ 7 2 π = 21 25 π ( M ) atau k = 4 → x = 21 2 π + 4 ⋅ 7 2 π = 21 26 π ( M ) k = 5 → x = 21 1 π + 5 ⋅ 7 2 π = 21 31 π ( M ) atau k = 5 → x = 21 2 π + 5 ⋅ 7 2 π = 21 32 π ( M ) k = 6 → x = 21 1 π + 6 ⋅ 7 2 π = 21 37 π ( M ) atau k = 5 → x = 21 2 π + 5 ⋅ 7 2 π = 21 38 π ( M ) Keterangan: M (Memenuhi) Dengan demikian, himpunan penyelesaiannya adalah .

Jawaban yang benar untuk pertanyaan tersebut adalah ${\frac{1}{21} π, \frac{2}{21} π, \frac{7}{21} π, \frac{8}{21} π, \frac{13}{21} π, \frac{14}{21} π, \frac{19}{21} π, \frac{20}{21} π, \frac{25}{21} π, \frac{26}{21} π, \frac{31}{21} π, \frac{32}{21} π, \frac{37}{21} π, \frac{38}{21} π}$ .

Persamaan Trigonometri Dasar

Ingat persamaan trigonometri dasar berikut:

Jika $sin x = sin α$ , nilai $x = α + k \cdot 2 π$ atau $x = (π - α) + k \cdot 2 π$ .

$sin 7 x = sin \frac{1}{3} π, 0 \leq x \leq 2 π$

$sin 7 x 7 x x = = = sin \frac{1}{3} π \frac{1}{3} π + k \cdot 2 π \frac{1}{21} π + k \cdot \frac{2}{7} π$

atau

$sin 7 x 7 x 7 x x = = = = sin \frac{1}{3} π (π - \frac{1}{3} π) + k \cdot 2 π \frac{2}{3} π + k \cdot 2 π \frac{2}{21} π + k \cdot \frac{2}{7} π$

Diperoleh:

$k = 0 \to x = \frac{1}{21} π + 0 \cdot \frac{2}{7} π = \frac{1}{21} π (M) atau k = 0 \to x = \frac{2}{21} π + 0 \cdot \frac{2}{7} π = \frac{2}{21} π (M) k = 1 \to x = \frac{1}{21} π + 1 \cdot \frac{2}{7} π = \frac{7}{21} π (M) atau k = 1 \to x = \frac{2}{21} π + 1 \cdot \frac{2}{7} π = \frac{8}{21} π (M) k = 2 \to x = \frac{1}{21} π + 2 \cdot \frac{2}{7} π = \frac{13}{21} π (M) atau k = 2 \to x = \frac{2}{21} π + 2 \cdot \frac{2}{7} π = \frac{14}{21} π (M) k = 3 \to x = \frac{1}{21} π + 3 \cdot \frac{2}{7} π = \frac{19}{21} π (M) atau k = 3 \to x = \frac{2}{21} π + 3 \cdot \frac{2}{7} π = \frac{20}{21} π (M) k = 4 \to x = \frac{1}{21} π + 4 \cdot \frac{2}{7} π = \frac{25}{21} π (M) atau k = 4 \to x = \frac{2}{21} π + 4 \cdot \frac{2}{7} π = \frac{26}{21} π (M) k = 5 \to x = \frac{1}{21} π + 5 \cdot \frac{2}{7} π = \frac{31}{21} π (M) atau k = 5 \to x = \frac{2}{21} π + 5 \cdot \frac{2}{7} π = \frac{32}{21} π (M) k = 6 \to x = \frac{1}{21} π + 6 \cdot \frac{2}{7} π = \frac{37}{21} π (M) atau k = 5 \to x = \frac{2}{21} π + 5 \cdot \frac{2}{7} π = \frac{38}{21} π (M)$