Giải PTLG sau KHÓ 3:

Question

$1) 4cos^3x+3\sqrt{2}sin2x=8cosx$ ;

$2) 2sin3x-\frac{1}{sinx}=2cos3x+\frac{1}{cosx}$ .

score 1 · Answer 1

Câu 1.

$4\cos^3 x+6\sqrt 2\sin x \cos x-8\cos x=0$

$\Leftrightarrow \cos x (2\cos^2 x +3\sqrt 2 \sin x -4)=0$ dễ rồi

Chú ý

$2\cos^2 x +3\sqrt 2 \sin x -4=0$

$\Leftrightarrow 2(1-\sin^2 x)+3\sqrt 2 \sin x -4=0$ tự làm

Câu 2 Điều kiện

$\sin 2x \ne 0$ tự làm

PT

$\Leftrightarrow 2\sin 3x \sin x \cos x-\cos x= 2\cos 3x \sin x \cos x +\sin x$

$\Leftrightarrow \sin 3x \sin 2x -\cos 3x \sin 2x -\sin x -\cos x=0$

$\Leftrightarrow \sin 2x (\sin 3x -\cos 3x) -(\sin x +\cos x)=0$

$\Leftrightarrow \sin 2x (3\sin x -4\sin^3 x -4\cos^3 x +3\cos x)-(\sin x +\cos x)=0$

$\Leftrightarrow \sin 2x \bigg [ 3(\sin x+\cos x) -4(\sin x +\cos x)(1-\sin x \cos x )\bigg ]-(\sin x +\cos x)=0$

+

$\sin x +\cos x = 0$ tự làm

+

$\sin 2x (3-4+4\sin x \cos x) -1=0$

$\Leftrightarrow \sin 2x (-1+2\sin 2x)-1=0$ tự làm

1 Đáp án