KST

1:  $2\sin (3x+\frac{\pi }{4}) = \sqrt{1+8.\sin 2x . \cos^{2} 2x}$
2:  $\cos 9x + 2\cos (6x+\frac{2\pi }{3}) +2 =0$
3:  $2.(1+\sin 2x) = tan(\frac{\pi }{4}+x)$
$đặt : a =3x+\frac{\pi }{3}$
$\Rightarrow pt :cos(3u-\pi)+2cos2u+2=0$
$\Leftrightarrow cos3u+2(2cos^2-1)+2=0$
$\Leftrightarrow 3cosu-4cos^3u+4cos^2u=0$
$\Leftrightarrow4cos^3u-4cos^2u-3cosu=0$
$\Leftrightarrow cosu(4cos^2u-4cosu-3)=0$ 
Đến đây chia TH tìm được $u\Rightarrow x$
Bạn lm nốt nhé
1)
bình phương 2 vế pt ta được $4sin^2(3x+\frac{\pi }{4})=1+8sin2x.cos^2 2x$
$\Leftrightarrow 2\left[ {} \right.1-cos(6x+\frac{\pi }{2})]=1+4sin2x.cos2x$ (hạ bậc $4sin^2x(...) nha )$
$\Leftrightarrow 2(1+sin6x)=1+2sin6x+2sin2x$
$\Leftrightarrow2+2sin6x=1+2sin6x+2sin2x$
$\Leftrightarrow sin2x=\frac{1}{2}$
bạn giải nốt nhé

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