^o^"懂"就是聰明Maths (2)

2010-09-20 2:46 am

Solve these 2 equations for 0<x<360

3b) 2 cot x = tan x +1

3g) cos x + sec x = 2

* show full steps, can u explain it and tell me the skills in doing??

更新1:

the ans for the first one is wrong it should be 45, 225, 116.6 , 251.6

回答 (2)

[匿名]

2010-09-20 3:14 am

✔ 最佳答案

1)
2cotx = tanx + 1

2cosx / sinx = sinx / cosx + 1

2 / tanx = tanx + 1

Mutiply by tanx,

2 = tan²x + tanx

tan²x + tanx - 2 = 0

(tanx + 2)(tanx - 1) = 0

tanx = -2 or tanx = 1

x = -63.4 or 117 or 45 or 225

So, the solutions are -63.4, 45, 117, 225

2)
cosx + secx = 2

cosx + 1 / cosx = 2

Mutiply by cosx

cos²x + 1 = 2cosx

cos²x - 2cosx + 1 = 0

(cosx - 1)² = 0

cosx = 1

x = 0, 360

So, the solutions are 0 and 360.

2010-09-19 20:17:29 補充：
ok.......

1)
2cotx = tanx + 1

2 / tanx = tanx + 1

2 = tan²x + tanx

tan²x + tanx - 2 = 0

(tanx + 2)(tanx - 1) = 0

tanx = -2 or tanx = 1

x = 45, 225, 116.6, 251.6

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Keiis

2010-09-20 3:18 am

the ans for the first one is wrong it should be 45, 225, 116.6 , 251.6

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