Tolong bantu nomer 3

f(x) = cosec^2 x - cot x cosec x
= 1/(sin x)^2 - (cos x)/(sin x) . 1/(sin x)
= 1/(sin x)^2 - (cos x)/(sin x)^2
= (1 - cos x)/(sin^2 x)
= (1 - cos x)/(1 - cos^2 x)
= (1 - cos x)/(1 - cos x)(1 + cos x)
= 1/(1 + cos x)
= (1 + cos x)^-1
f(x) turun jika f'(x) < 0
f'(x) = -1(1 + cos x)^-2 . (-sin x) < 0
=> (sin x) / (1 + cos x)^2 < 0
=> sin x = 0 atau cos x = -1
1) sin x = 0 => x = 0°, 180°, 360°
2) cos x = -1 => x = 180°
Garis bilangan
----- (0°) ++++ (180°) ---- (360°) +++
x < 0° atau 180° < x < 360°
Karena interval nya 0 ≤ x ≤ 360° maka yang memenuhi
180° < x < 360°