2cos^2θ + sinθcosθ-sin^2θ = 0
Thanks!!! :)
Maths 題！！
2011-01-24 5:33 am
回答 (2)
2011-01-24 5:44 am
✔ 最佳答案
2cos^2θ+sinθcosθ-sin^2θ = 0(2cosθ-sinθ)(cosθ+sinθ)=0
2cosθ-sinθ=0 or cosθ+sinθ=0
sinθ=2cosθ or sinθ=-cosθ
tanθ=2 or tanθ=-1
θ=180°n+63.4° or θ=180°n-45° (where n is an integer)
2011-01-24 5:40 am
What is the range of θ?
Or required to solve general solution?
Or required to solve general solution?
收錄日期: 2021-04-22 22:18:02
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110123000051KK01295