Giúp mình nữa nha

T

tettrungthu17896

0

01697981687

câu4: 25sin2x-4sinx+3cosx-9=0
<=>50sinxcosx-4sinx+3cosx-9=0
<-=>50\frac{2tan(x/2)}{1+tan^2(x/2)}\frac{1-tan^2(x/2)}{1+tan^2(x/2)}-4\frac{2tan(x/2)}{1+tan^2(x/2)}+3\frac{1-tan^2(x/2)}{1+tan^2(x/2)}-9=0
<=>100tan(x/2)(1-tan^2(x/2))-8tan(x/2)(1+tan^2(x/2))+3(1-tan^4(x/2))-9(1+tan^2(x/2))^2=0
<=>12tan^4(x/2)+tan^3(x/2)+118tan^2(x/2)-92tan(x/2)+6=0
mình giải thế có gì sai sữa giùm

câu2: $16sin^3xcos^2(2x)+sin7x=12+3sin5x$
$<=>4(3sinx-sin3x)cos^2(2x)+sin7x=12+3sin5x$
$<=>12sinxcos^2(2x)-4sin3xcos^2(2x)+sin7x=12+3sin5x$
$<=>6cos2x(sin3x-sinx)-2cos2x(sin5x+sinx)+sin7x=12+3sin5x$
$<=>6cos2xsin3x-6cos2xsinx-2sin5xcos2x-2sinxcos2x+sin7x=12+3sin5x$
$<=>3(sin5x+sinx)-3(sin3x-sinx)-(sin7x+sin3x)-(sin3x-sinx)+sin7x=12+3sin5x$
$<=>5sin3x-7sinx+12=0$
mình giải như thế có sai xót gì sữa giùm mình nha tk nh`

Câu 1:
$cosx+\sqrt{3}(sin2x+sinx)-4cos2xcosx-2cos^2x+2=0$
$<=>cosx+2\sqrt{3}sinxcosx+\sqrt{3}sinx-2(cos3x+cosx)-2cos^2x+2=0$
$<=>2\sqrt{3}sinxcosx+\sqrt{3}sinx-2cos3x-cosx-2cos^2x+2=0$
$<=>\sqrt{3}sinx(2cosx+1)-cosx(1+2cosx)+2(1-cos3x)=0$
$<=>\sqrt{3}sinx(2cosx+1)-cosx(1+2cosx)-2(4cos^3x-3cosx-1)=0$
$<=>\sqrt{3}sinx(2cosx+1)-cosx(1+2cosx)-2(cosx-1)(2cosx+1)^2=0$
$<=>(2cosx+1)(\sqrt{3}sinx-cosx-4cos^2x+cosx+1)=0$
$<=>(2cosx+1)(\sqrt{3}sinx-4cos^2x+1)=0$
đến đây tự giải nhé
 
Last edited by a moderator:
You must log in or register to reply here.
Top Bottom