Solve cos^2(x)-sin^2(x)=2sin(2x), which x is between 0 deg and 360 deg.
The answer is : 13.3 deg ; 103.3 deg ; 193.3 deg ; 283.3 deg.
How to get these answers? Please help!
Thank you !
trigo problem
2010-01-21 5:23 am
回答 (3)
2010-01-21 5:41 am
✔ 最佳答案
cos^2(x)-sin^2(x)=2sin(2x)cos^2(x)-sin^2(x)-2sin(2x)=0
cos^2(x)-sin^2(x)-4sinxcosx=0
[cos^2(x)-sin^2(x)-4sinxcosx]/cos^2(x)=0
1-tan^2(x)-4tanx=0
tan^2(x)+4tanx-1=0
tanx= [-4+2(5)^(1/2)]/2 or [-4-2(5)^(1/2)]/2
tanx=0.236 or -4.236
x=13.3 deg or 193.3 deg or 103.3 deg or 283.3 deg
2010-01-21 6:45 am
how can tan2x = 1/2 get to 2x = 26.6, 206.6, 386.6, 566.6?
2010-01-21 6:33 am
cos^2x - sin^2x = 2sin2x
cos2x = 2sin2x
tan2x = 1/2
2x = 26.6, 206.6, 386.6, 566.6
x = 13.3, 103.3, 193.3, 283.3
2010-01-20 23:10:11 補充：
tan2x = 1/2 => 2x = 26.6, 26.6+180,26.6+360, 26.6+540
即26.6 + 180的倍數
cos2x = 2sin2x
tan2x = 1/2
2x = 26.6, 206.6, 386.6, 566.6
x = 13.3, 103.3, 193.3, 283.3
2010-01-20 23:10:11 補充：
tan2x = 1/2 => 2x = 26.6, 26.6+180,26.6+360, 26.6+540
即26.6 + 180的倍數
收錄日期: 2021-04-23 23:20:35
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