amath..trigo!急,,20pts－kfsdgte的部落格

標題:

amath..trigo!急,,20pts

發問:

solved the following equations: * 0
最佳解答:

1) sec y tan y - tany +sec y -1 =0 secy(tany+1)-(tany+1)=0 (tany+1)(secy-1)=0 tany=-1 or secy=1 tany=-1 or cosy=1 y=0,135, 315, 360 2) sin^4x-cos^4x= cos x -cosx=cos^4x-sin^4x -cosx=(cos^2x-sin^2x)(cos^2x+sin^2x) -cosx=cos^2x-sin^2x -cosx=cos2x -cosx=2cos^2x-1 2cos^2x+cosx-1=0 (2cosx-1)(cosx+1)=0 cosx=1/2 or cosx=-1 x=60,180 or 300 3) If the equation 2x^2 - (4cosx)y+3sinx =0 in y has equal roots,find the values of x between 0 and 360. Since the equation has equation root discriminant=0 (4cosx)^2-24sinx=0 16cos^2x-24sinx=0 16(1-sin^2x)-24sinx=0 16sin^2x+24sinx-16=0 2sin^2x+3sinx-2=0 (2sinx-1)(sinx+2)=0 sinx=1/2 or -2 (rejected) so x =30 ,150 2007-01-30 00:23:15 補充： Note that in (1) 0, 360 are not rejected 2007-01-30 00:27:22 補充： 3 應是has equal root

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