Solve the equation 3cosA+4sinA=2
回答 (1)
2007-05-12 1:37 pm
✔ 最佳答案
3cosA+4sinA=2 (3/5)cosA+(4/5)sinA=2/5
cosBcosA+sinBsinA=2/5 ( where cos B=3/5, sin B=4/5)
cos(A-B)=2/5
cos(A-53.13)=2/5
A-53.13=360n+66.42 or 360n-66.42 (n is an integer)
A=360n+119.55 or 360n-13.29 (n is an integer)
收錄日期: 2021-04-12 21:10:56
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070512000051KK00459