Can you please help me
4sinx - 2cosx = 3
required to find the value of x, where 0<=x<=360.
thank you
S4 Maths Trigonometry
2010-04-24 6:59 am
回答 (2)
2010-04-24 7:14 am
✔ 最佳答案
=====================================================圖片參考：http://img231.imageshack.us/img231/1036/77287975.png
2010-04-23 23:14:42 補充：
http://img231.imageshack.us/img231/1036/77287975.png
2010-04-24 7:22 am
Sol
4Sinx－2Cosx=3
√20(4/√20Sinx－2/√20Cosx)=3
4/√20Sinx－2/√20Cosx=3/√20
Set Cosw=2/√5，sSnw=1/√5，0<90度
Tanw=1/2，w=26.565度
CoswSinx－SinwCosx=3/√20
Sin(x－w)= 3/√20
Sin(x－26.565度)= 0.67082
0<=x<=360度
－26.565度<=x－26.565度<=333.435度
x－26.565度=42.13度 or x－26.565度=137.87度
x=68.695度 or x=164.435度
4Sinx－2Cosx=3
√20(4/√20Sinx－2/√20Cosx)=3
4/√20Sinx－2/√20Cosx=3/√20
Set Cosw=2/√5，sSnw=1/√5，0<90度
Tanw=1/2，w=26.565度
CoswSinx－SinwCosx=3/√20
Sin(x－w)= 3/√20
Sin(x－26.565度)= 0.67082
0<=x<=360度
－26.565度<=x－26.565度<=333.435度
x－26.565度=42.13度 or x－26.565度=137.87度
x=68.695度 or x=164.435度
收錄日期: 2021-04-23 23:21:40
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100423000051KK01685