1/(1+tan(3θ))=3^0.5
θ=?
3^0.5是指開方3
三角學的方程式
2010-06-13 7:36 am
回答 (3)
2010-06-13 4:11 pm
第一位漏左部份答案... tan3x 應該有6個解先岩
2010-06-13 7:51 am
2010-06-13 7:49 am
1/(1+tan(3θ))=3^0.5
1/開方3 = 1+ tan 3θ
tan 3θ = 1/開方3 -1
3θ = -22.911 or 157.08866
θ = -7.637 or 52.363
θ = 52.363
2010-06-14 23:51:59 補充：
我知要幾多range 喎
順便加多個lo
=======================================
157.08866 +180
= 337.08866
1/開方3 -1 = tan 337.08866
θ =112.363
======================================
157.08866 +360
= 517.08866
1/開方3 -1 = tan 517.08866
θ =172.363
======================================
θ =232.363
2010-06-14 23:53:27 補充：
θ =292.363
θ =352.363
θ =412.363
.............. 無限
1/開方3 = 1+ tan 3θ
tan 3θ = 1/開方3 -1
3θ = -22.911 or 157.08866
θ = -7.637 or 52.363
θ = 52.363
2010-06-14 23:51:59 補充：
我知要幾多range 喎
順便加多個lo
=======================================
157.08866 +180
= 337.08866
1/開方3 -1 = tan 337.08866
θ =112.363
======================================
157.08866 +360
= 517.08866
1/開方3 -1 = tan 517.08866
θ =172.363
======================================
θ =232.363
2010-06-14 23:53:27 補充：
θ =292.363
θ =352.363
θ =412.363
.............. 無限
收錄日期: 2021-04-25 17:01:04
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100612000051KK01802