F.5 TRIGONOMETRY

2012-10-22 9:04 am
solve the following equations for 0°< x< 360°

1)4sin^2x-9sinxcosx+2cos^2x=0

2)tan^2(2x-10°)=3

3)2sin2x+1=0

4)tan^2-tanx-6=0

回答 (2)

2012-10-22 9:59 am
✔ 最佳答案
1)
4 sin² x - 9 sin x cos x + 2 cos² x = 0
(4 sin² x - 9 sin x cos x + 2 cos² x) /cos² x = 0
4 tan² x - 9 tan x + 2 = 0
(4 tan x - 1)(tan x - 2) = 0
tan x = 1/4 or tan x = 2
x = 14.04°, (180+14.04)° or x = 63.43°,(180+63.43)°
x = 14.04°, 63.43°,194.04°, 243.43°


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2)
0° < x < 360°
-10° < (2x - 10°) < 710°

tan² (2x - 10°) = 3
tan (2x - 10°) = √3 或tan (2x - 10°) = -√3
(2x - 10°) = 60°,(180+60)°, (360+60)°, (540+60)° or (2x - 10°) = (180-60)°, (360-60)°, (540-60)°, (720-60)°
x = 35°, 65°, 125°, 155°, 215°, 245°, 305°,335°


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3)
0° < x < 360°
0° < 2x < 720°

2 sin 2x + 1 = 0
2 sin 2x = -1
sin 2x = -1/2
2x = (180+30)°, (360-30)°,(540+30)°, (720-30)°
x = 105°, 165°, 285°, 345°


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4)
tan² x- tan x - 6 = 0
(tan x - 3)(tan x + 2) = 0
tanx = 3 or tan x = -2
x = 71.57°,(180+71.57)° or x = (180-63.43)°, (360-63.43)°
x = 71.57°, 116.57°, 251.57°, 296.57°
參考: 土扁
2013-05-25 6:43 pm
呢d都叫F5題目,你是但找個中四學生都識做啦


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