標題:

F.5 math

發問:

1(a) Factorize 8x^2 - 6xy - 9y^2. (b) Hence solve 8sin^2 x - 9cos^2 x = 6sinxcosx for 0<=x<=360. 2(a) Solve 2sin^2 x+ sinxcosx - cos^2 x = 0 for 0<=x<=360. (b) Hence solve 1/ (sin^2 x) = 1/(tanx) + 3 for 0<=x<=360.

最佳解答:

1(a) Factorize 8x^2 - 6xy - 9y^2. 8x^2-6xy-9y^2 =(2x-3y)(4x+3y) (b) Hence solve 8sin^2 x - 9cos^2 x = 6sinxcosx for 0<=x<=360. 8(sinx)^2-9(cosx)^2=6(sinx)(cosx) 8(sinx)^2-6(cosx)(sinx)-9(cosx)^2=0 by (a),put x=sinx,y=cosx, [2(sinx)-3(cosx)][4(sinx)+3(cosx)]=0 2(sinx)-3(cosx)=0 or 4(sinx)+3(cosx)=0 2(sinx)=3(cosx) or 4(sinx)=-3(cosx) tanx=3/2 or tanx=-3/4 x=56.30993247 degrees or x=236.3099325 degrees or x=143.1301024 degrees or x=323.1301024 degrees 2(a) Solve 2sin^2 x+ sinxcosx - cos^2 x = 0 for 0<=x<=360. [2(sinx)-(cosx)][(sinx)+(cosx)]=0 2(sinx)-(cosx)=0 or (sinx)+(cosx)=0 tanx=1/2 or tanx=-1 x=26.56505118 degrees or x=206.5650512 degrees or x=135 degrees or 315 degrees (b) Hence solve 1/ (sin^2 x) = 1/(tanx) + 3 for 0<=x<=360. 1/[(sinx)^2]=1/(tanx)+3 1/[(sinx)^2]-1/(tanx)-3=0 [1-(sinx)(cosx)-3(sinx)^2]/[(sinx)^2]=0 if x=0 degrees ,180 degrees, sinx=0 there is no solutions for x=/=0 degrees and x=/=180 degrees, 1-(sinx)(cosx)-3(sinx)^2=0 since 1-(sinx)^2=(cosx)^2, (cosx)^2-(sinx)(cosx)-2(sinx)^2=0 2(sinx)^2+(sinx)(cosx)-(cosx)^2=0 by(a), x=26.56505118 degrees or x=206.5650512 degrees or x=135 degrees or 315 degrees

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