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°
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2) cos x = -1 => x = 180° Kenapa ada cos x = -1
Verified answer
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°