bantu dong nomer 3 dan 4

Mapel : Matematika
Kelas : X SMA
Bab : Logaritma dan Trigonometri

Pembahasan :
3.

4√[x - 5] < x - 1
16(x - 5) < (x - 1)²
16x - 80 < x² - 2x + 1
x² - 18x + 81 > 0
(x - 9)² > 0
x < -9 atau x > 9

Syarat
x - 5 ≥ 0
x ≥ 5

Gabungkan Hp nya...
Hp : {x > 9}

P = x - 12
P = {-2 , -1 , 0 , ...}
P = x ≥ -3, x ≠ -3

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4.

Tan x = -3/4
Sin x = -3/5
Cos x = 4/5

Sin (π/3 - x)
= Sin (60 - x)
= Sin 60 Cos x - Cos 60 Sin x
= 1/2 √3 (4/5) - (1/2)(-3/5)
= 2/5 √3 + 3/10
= (4√3 + 3)/10

Jawaban : C

3) p = x - 12 => x = p + 12
4√(x - 5) < x - 1
4√(p + 12 - 5) < p + 12 - 1
4√(p + 7) < p + 11 ====> kedua ruas dikuadratkan
(4√(p + 7))^2 < (p + 11)^2
16(p + 7) < p^2 + 22p + 121
16p + 112 < p^2 + 22p + 121
-p^2 - 6p - 9 < 0
p^2 + 6p + 9 > 0
(p + 3)(p + 3) > 0
p = -3 atau p = -3
+++++ (-3) +++++
p ≠ -3
Syarat : √(p + 7) => p + 7 ≥ 0 => p ≥ -7
HP : p ≥ -7, p ≠ -3

4) 3π/2 ≤ x ≤ 2π => dikuadran 4 (cos = (+), Tan = (-) dan sin = (-))
Tan x = -3/4 = de/sa
mi = √(3^2 + 4^2) = √(9 + 16) = √25 = 5
sin x = -de/mi = -3/5
cos x = sa/mi = 4/5
sin (π/3 - x) = sin π/3 cos x - cos π/3 sin x
= 1/2 √3 . 4/5 - 1/2 . -3/5
= 4√3/10 + 3/10
= (4√3 + 3)/10