giup t 2 bai luong giac

N

nhoccon_sieuquay94

a/ 32cos^6(x)-cos6x=4
<=>4(1+cos2X)^3 - 4cos^3(2x)+3cos(2x)-4=0
<=>4(1+3cos2x+3cos^2(2x)+cos^3(2x))-4cos^3(2x)+3cos2x-4=0
<=>12cos^2(2x)+15cos2x=0
<=>cos2x=0 hoac cos2x=-15/12..........ban tu giai? tiep' nha
b/cos3x=cos^2(9x/4)
<=>4cos^2(3x/2)-2=1+cos(9x/2)
<=>4coss^2(3x/2)-3-4cos^3(3x/2)+3cos(3x/2)=0
<=>4cos^3(3x/2)-4cos^2(3x/2)-3cos(3x/2)+3=0
<=>cos(3x/2)=1 (*) hoac 4cos^2(3x/2)-3= 0 (**)
giai? pt (**)
ta co
2(1+cos3x)-3=0 <=>cos3x=1/2 tu giai? tiep' nha
 
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