Đáp án:
Giải thích các bước giải:
$cos3x.cos4x+sin2x.sin5x=`1/2`.cos2x+cos4x$
$<=>Cos4x.cos3x+sin5x.sin2x=1/2.(cos4x+cos2x)$
$<=>1/2(cos7x+cosx)-1/2(cos7x+cos3x)=cos3xcosx$
$<=>cos3x+cosx-2cos3xcosx=0$
$<=>2cos2xcosx-2cos3xcosx=0$
$<=>cosx(cos2x-cos3x)=0$
4,
<=> cos5x. cosx = cos4x. Cos2x +3cos^2x +1
<=> `1/2`. (cos6x + cos4x) =`1/2`. (cos6x +cos2x) +3cos^2x +1
<=> cos4x = cos2x +6cos^2x +2
<=> 2cos^2( 2x )-4 cos2x - 6=0
<=> cos2x =-1 or cos2x =3 (PTVN)
