12) (tan^2 x)/(1 + sec x) = 1 tan^2 x = 1 + sec x (sin^2 x)/(cos^2 x) = 1 + 1/(cos x) ====> kedua ruas dikali cos^2 x sin^2 x = cos^2 x + cos x (1 - cos^2 x) = cos^2 x + cos x 0 = 2 cos^2 x + cos x - 1 Misal cos x = p 2p^2 + p - 1 = 0 (2p - 1)(p + 1) = 0 p = 1/2 atau p = -1 cos x = 1/2 => cos x = cos 60° = cos 300° cos x = -1 => cos x = cos 180° Karena 0 < x < 90 maka x = 60°
13) cos 5π/6 = cos 5(180°)/6 = cos 150° = - cos 30° = -1/2 √3
Verified answer
12) (tan^2 x)/(1 + sec x) = 1tan^2 x = 1 + sec x
(sin^2 x)/(cos^2 x) = 1 + 1/(cos x) ====> kedua ruas dikali cos^2 x
sin^2 x = cos^2 x + cos x
(1 - cos^2 x) = cos^2 x + cos x
0 = 2 cos^2 x + cos x - 1
Misal cos x = p
2p^2 + p - 1 = 0
(2p - 1)(p + 1) = 0
p = 1/2 atau p = -1
cos x = 1/2 => cos x = cos 60° = cos 300°
cos x = -1 => cos x = cos 180°
Karena 0 < x < 90 maka x = 60°
13) cos 5π/6 = cos 5(180°)/6 = cos 150° = - cos 30° = -1/2 √3