easy a.math

2007-04-12 12:56 am
pls solve this equation 0<=x<=360 degree
cosx+cos2x+cos3x+cos4x=0

回答 (3)

2007-04-12 1:31 am
參考: My Maths knowledge
2007-04-12 1:45 am
hi,雷欣同學,現在為你解答:

cos x +cos 2x + cos 3x + cos 4x =0
cos x + cos 3x + cos 2x +cos 4x =0
2cos 2xcos x + 2cos3xcosx =0(cosA+cosB=2{cos(A+B)/2}{cos(A-B)/2})(呢步ok既,siu sir唔要求咁多)
cos 2x + cos 3x =0
2cos(5x/2)cos(x/2)=0
so cos (5x/2)=0 or cos (x/2)=0 or cos x=0
5x/2=90,270,450,810 or x/2=90 or x=90,270,
x=36,108,180,252,324 or x=180

so, x=36,190,108,180,252,270,324(degree 自己加)
2007-04-12 1:34 am
cos x + cos 2x + cos 3x + cos 4x = 0
(cos x + cos 3x) + (cos 2x + cos 4x) = 0
2(cos 2x)(cos x) + 2(cos 3x)(cos x) = 0
(cos x)(cos 2x + cos 3x) = 0
cos x = 0 or 2(cos 2.5x)(cos 0.5x) = 0
cos x = 0 or cos 2.5x = 0 or cos 0.5x = 0
(x = 90 or 270 degrees) or (2.5x = 90, 270, 450, 630 or 810 degrees) or (0.5x = 90 degrees)
x=36, 90, 108, 180, 252, 270 or 324 degrees


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