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secondary 3 | A Maths
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can someone pls help me solve these 3 equations I can’t seem to get the correct answer
Date Posted: 2 years ago
Views: 263
40 = 35 • sin² x - 54 • sin x • cos x
40 • sin² x + 40 • cos² x
= 35 • sin² x - 54 • sin x • cos x
5 • sin² x + 54 • sin x • cos x + 40 • cos² x = 0
(sin x+10cos x) • (5sin x+4cos x) = 0
sin x + 10cos x = 0
tan x = -10
x = 275.71°+360° = 635.71°
5sin x + 4cos x = 0
tan x = -0.8
x = 321.34° + 360° = 681.34°
40 • sin² x + 40 • cos² x
= 35 • sin² x - 54 • sin x • cos x
5 • sin² x + 54 • sin x • cos x + 40 • cos² x = 0
(sin x+10cos x) • (5sin x+4cos x) = 0
sin x + 10cos x = 0
tan x = -10
x = 275.71°+360° = 635.71°
5sin x + 4cos x = 0
tan x = -0.8
x = 321.34° + 360° = 681.34°
2•sin x•cos x = sin² x + cos² x
sin² x - 2•sin x•cos x + cos² x = 0
(sin x - cos x)² = 0
sin x = cos x
x₁ = 45°+360° = 405°
x₂ = 225° ✘
sin² x - 2•sin x•cos x + cos² x = 0
(sin x - cos x)² = 0
sin x = cos x
x₁ = 45°+360° = 405°
x₂ = 225° ✘
9•sin² 2x - 19 cos 2x + 11 • cos ² 2x = 0
9 - 19 cos 2x + 2• cos ² 2x = 0
(2•cos 2x - 1)•(cos 2x - 9) = 0
2•cos 2x - 1 = 0
cos 2x = 0.5
x= 30° （Answer）
cos 2x - 9 = 0
cos 2x = 9 ✘
9 - 19 cos 2x + 2• cos ² 2x = 0
(2•cos 2x - 1)•(cos 2x - 9) = 0
2•cos 2x - 1 = 0
cos 2x = 0.5
x= 30° （Answer）
cos 2x - 9 = 0
cos 2x = 9 ✘
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Part a
Date Posted:
2 years ago
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Second and third part
Date Posted:
2 years ago
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