A.Sin 2x=sin ¾ phi,0°<0<2phi B.Tan x= tan 126°,0° Tolong dengan cara pengerjaan nya, makasih..
kekkaiisshi
Sin 2x = sin 3/4 π 2x = 3/4π + k.2π x = 3/8π + k.π k = 0, x = 3/8π k = 1, x = 11/8 π atau 2x = (π - 3/4π) + k.2π 2x = 1/4π + k.2π x = 1/8π + k.π k = 0, x = 1/8π k = 1, x = 9/8π
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Takamori37
Melengkapi yang bawah: tan x = tan 126° Dengan: 0 ≤ x ≤ 360° Maka, Solusi tangen: tan x = tan a Didapat, x = a + k.180° Maka, Solusi x = 126 + k.180° Dengan k = {0,1} Diperoleh: x = 126° x = 306°
2x = 3/4π + k.2π
x = 3/8π + k.π
k = 0, x = 3/8π
k = 1, x = 11/8 π
atau
2x = (π - 3/4π) + k.2π
2x = 1/4π + k.2π
x = 1/8π + k.π
k = 0, x = 1/8π
k = 1, x = 9/8π
tan x = tan 126°
Dengan:
0 ≤ x ≤ 360°
Maka,
Solusi tangen:
tan x = tan a
Didapat,
x = a + k.180°
Maka,
Solusi
x = 126 + k.180°
Dengan k = {0,1}
Diperoleh:
x = 126°
x = 306°