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Trigonometri

2) Hp dari  2 sin 3x - √2 = 0 , pada 0≤ x ≤ 360
.
2 sin 3x  = √2
sin 3x = ¹/₂√2 = sin 45
3x = 45 +k. 360 atau 3x = (180-45) + k. 360
x = 15 + k. 120  atau x = 45 + k.120
k  = 0,1,2,3,...
k = 0-->  x₁ = 15, x₂ = 45
k=1, --> x₃ = 135, x₄ = 165
k = 2 --> x₅ = 255, x₆ = 285
k = 3 --> x₇ = 375 (tm) , x₈= 405 (tm)

HP x ={14,45,135,165,255,285}


4a) sin A/(1-cos A) = (1+cos A)/sin A
ruas kiri sin A/(1-cos A)  X (1+cos A)/(1+cos A)=
= {sinA(1 + cos A)} / {(1+cos A)(1-cos A)}
= {sin A (1+cos A) / (1 - cos²A)
= {sin A (1+cos A) / (sin² A)
= (1 + cos A)/(sin A) = ruas kanan --> terbukti

4b). sin A + cos A. cot A = cosec A
ruas kiri  sin A + cos A . cot A .....kalikan sin A/sin A
= {sin A + ( cos A . cos A /sin A)} (sin A /sin A)
= { sin A  sin A + (cos A. cos A)} / (sin A)
= (sin² A + cos² A) / sin A
= 1/sin A
= cosec A = ruas kanan (terbukti)