KST

Question

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: $2.(1+\sin 2x) = tan(\frac{\pi }{4}+x)$

Kaito kid · Answer 1 · 8/4/2016 11:32:44 AM

$đặt : a =3x+\frac{\pi }{3}$

$\Rightarrow pt :cos(3u-\pi)+2cos2u+2=0$

$\Leftrightarrow cos3u+2(2cos^2-1)+2=0$

$\Leftrightarrow 3cosu-4cos^3u+4cos^2u=0$

$\Leftrightarrow4cos^3u-4cos^2u-3cosu=0$

$\Leftrightarrow cosu(4cos^2u-4cosu-3)=0$

Đến đây chia TH tìm được $u\Rightarrow x$

Bạn lm nốt nhé

Trả lời 04-08-16 11:32 AM

Kaito kid
1

80K 161K

Kaito kid · Answer 2 · 8/3/2016 4:59:31 PM

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

Trả lời 03-08-16 04:59 PM

Kaito kid
1

80K 161K

2 Đáp án