Dah kelar ..... baru Unfoll ygy...
oke dah ....
3sin(x) = √3cos(x) Nilai x = ...
DIKETAHUI :
[tex] \sf \: 3 \sin(x) = \sqrt{3} \cos(x) [/tex]
DITANYA :
nilai X
JAWAB :
Langkah pertama
[tex] \sf \: \cos(x) = 0[/tex]
Selesaikan penyederhanaan dari X
[tex] \sf \: x = \frac{\pi}{2} + k\pi, k \in\mathbb{Z}[/tex]
[tex] \sf \: \frac{3 \sin(x) }{ \cos(x) } = \sqrt{3} \: \bf \: dimana \sf \: x \ne\frac{\pi}{2} + k\pi, k \in\mathbb{Z}[/tex]
[tex]\sf 3 \times \frac{\sin(x) }{ \cos(x) } = \sqrt{3} \: \bf \: dimana \sf \: x \ne\frac{\pi}{2} + k\pi, k \in\mathbb{Z}[/tex]
[tex] \sf \: \frac{ \sin(t) }{ \cos(t) } = \tan(t) [/tex]
[tex] \sf \: 3 \tan(x) = \sqrt{3} [/tex]
Langkah kedua
[tex] \sf \: 3 \tan(x) \div 3 = \sqrt{3} \div 3[/tex]
[tex] \sf \: \tan(x) = \frac{ \sqrt{3} }{3} [/tex]
Gunakan sudut lingkaran positif terkecil untuk mencari tan(x)
[tex] \sf \: x = \frac{\pi}{6} [/tex]
[tex] \sf \: x = \frac{\pi}{6} + k\pi, \: k \in\mathbb{Z}, \: x \ne\frac{\pi}{2} + k\pi, \: k \in\mathbb{Z} \\ \\ \sf \: x = \boxed{ \bf\frac{\pi}{6} + k\pi, k \in\mathbb{Z}}[/tex]
[answer.2.content]
