It is given that tan x=(1 + cos x) / (2 sin x)
(a) Show that 3 cos^2 x + cos x - 2 = 0
(b) Hence, solve tan x = (1 + cos x) / (2 sin x) for 0<x<360
(Give your answers correct to 1 decimal place if necessary.)
F.4 Mathematics (Trigonometry)
2013-03-25 1:41 am
回答 (3)
2013-03-25 2:22 am
✔ 最佳答案
(a)tan x = (1 + cos x) / (2 sin x)
sin x / cos x = (1 + cos x) / (2 sin x)
cos x (1 + cos x) = sin x (2 sin x)
cos x + cos² x = 2 sin² x
cos x + cos² x + 2 cos² x - 2 cos² x - 2 sin² x = 0
cos x + 3 cos² x - 2 (cos² x + sin² x) = 0
3 cos² x + cos x - 2 = 0
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(b)
tan x = (1 + cos x) / (2 sin x)
From (a) :
3 cos² x + cos x - 2 = 0
(3 cos x - 2)(cos x + 1) = 0
cos x = 2/3 or cos x = -1
x = 48.2° or x = (360-48.2)° or x = 180° (rejected)
x = 48.2° or x = 311.8°
(x = 180° is rejected because the denominator sin x should NOT be 0.)
參考: Adam
2013-03-25 2:22 am
tan x=(1 + cos x) / (2 sin x)
sin x / cos x = (1 + cos x) / (2 sin x)
[ sin x (2 sin x) ]/ cos x = (1 + cos x)
2 (sin x) ^2 = (1 + cos x) cos x
2[1 - (cos x )^2 ]= cos x + (cos x)^2
2 - 2(cos x )^2 = cos x + (cos x)^2
2=2(cos x )^2 +cos x + (cos x)^2
0= 3(cos x )^2 +cos x -2
3(cos x )^2 +cos x -2 =0
(B) tan x = (1 + cos x) / (2 sin x) for 0<x<360
tan x= (1 + cos x) / (2 sin x)
3(cos x )^2 +cos x -2 =0
(3cos x -2 )(cos x +1)=0
cos x =2/3 or cos x = -1
x = 48.2 ,180 ,311.8
sin x / cos x = (1 + cos x) / (2 sin x)
[ sin x (2 sin x) ]/ cos x = (1 + cos x)
2 (sin x) ^2 = (1 + cos x) cos x
2[1 - (cos x )^2 ]= cos x + (cos x)^2
2 - 2(cos x )^2 = cos x + (cos x)^2
2=2(cos x )^2 +cos x + (cos x)^2
0= 3(cos x )^2 +cos x -2
3(cos x )^2 +cos x -2 =0
(B) tan x = (1 + cos x) / (2 sin x) for 0<x<360
tan x= (1 + cos x) / (2 sin x)
3(cos x )^2 +cos x -2 =0
(3cos x -2 )(cos x +1)=0
cos x =2/3 or cos x = -1
x = 48.2 ,180 ,311.8
2013-03-25 2:00 am
3cos^2x+cosx-2
=(3cosx-2)(cosx+1)
then i don't know
=(3cosx-2)(cosx+1)
then i don't know
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