1. 2 cos (3x-30°)=√3 , 0≤x≤360° 2. tan (1/2x)= tan [tex] \pi [/tex] /6 0≤x≤2[tex] \pi [/tex] mohon bantuannya
Matematika
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Pertanyaan
1. 2 cos (3x-30°)=√3 , 0≤x≤360°
2. tan (1/2x)= tan [tex] \pi [/tex] /6 0≤x≤2[tex] \pi [/tex]
mohon bantuannya
2. tan (1/2x)= tan [tex] \pi [/tex] /6 0≤x≤2[tex] \pi [/tex]
mohon bantuannya
1 Jawaban
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1. Jawaban Sutr1sn0
1.
2 cos (3x - 30°) = akar 3 , 0<_x<_360°
cos (3x - 30°) = 1/2 (akar 3)
cos (3x - 30°) = cos 30°
3x - 30° = 30°
3x = 30° + 30°
3x = 60°
x = 60°/3
x = 20°
atau
2 cos (3x - 30°) = akar 3
cos (3x - 30°) = 1/2 (akar 3)
cos (3x - 30°) = cos 330°
3x - 30° = 330°
3x = 330° + 30°
3x = 360°
x = 360°/3
x = 120°
jadi hp nya adalah {20°,120°}
2.
tan (1/2 x) = tan phi/6 , 0<_x<_2phi
tan (1/2 x) = tan 180°/6
tan (1/2 x) = tan 30°
1/2 x = 30°
x = 30° × 2
x = 60°
x = 1/3 phi
atau
tan (1/2 x) = tan 210°
1/2 x = 210°
x = 210° × 2
x = 420°
x =/= 7/3 phi
jadi hp nya adalah {1/3 phi}