solve the following equations for 0<= x<= 360
2sin^2x - sin2x=2
a.maths ar
2007-02-21 2:11 am
回答 (1)
2007-02-21 2:34 am
✔ 最佳答案
2sin^2x-sin2x=22sin^2x-2sinxcosx-2=0
2sin^2x-2sinxcosx-2cos^2x-2sin^2x=0
cos^2x+sinxcosx=0
cosx(cosx+sinx)=0
cosx=0 or cosx=-sinx
x=90(rej) or 270 or 135 or 315
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收錄日期: 2021-04-12 18:01:11
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