Chủ đề này có 1 trả lời

[haobaibon123]

$tan(x-\frac{\pi}{6}).tan(x+\frac{\pi}{3})sin3x=sinx+sin2x$

[hoangtrong2305]

$tan(x-\frac{\pi}{6}).tan(x+\frac{\pi}{3})sin3x=sinx+sin2x$

$\tan(x-\frac{\pi}{6}).\tan(x+\frac{\pi}{3})\sin 3x=\sin x+\sin 2x$

ĐKXĐ:..................................................

$\tan(x-\frac{\pi}{6}).\tan(x+\frac{\pi}{3})\sin 3x=\sin x+\sin 2x$

$\Leftrightarrow \frac{\sin (x-\frac{\pi}{6}).\sin (x+\frac{\pi}{3})\sin 3x}{\cos (x-\frac{\pi}{6}).\cos(x+\frac{\pi}{3})}=\sin x+\sin 2x$

$\Leftrightarrow \sin (x-\frac{\pi}{6}).\sin (x+\frac{\pi}{3})\sin 3x=(\sin x+\sin 2x)\cos (x-\frac{\pi}{6}).\cos(x+\frac{\pi}{3})$

$\Leftrightarrow -2\sin (x-\frac{\pi}{6}).\sin (x+\frac{\pi}{3})\sin 3x=-2\cos (x-\frac{\pi}{6}).\cos(x+\frac{\pi}{3})(\sin x+\sin 2x)$

$\Leftrightarrow \cos (2x+\frac{\pi}{6}).\sin 3x=-\cos (2x+\frac{\pi}{6})(\sin x+\sin 2x)$

$\Leftrightarrow \cos (2x+\frac{\pi}{6}).\sin 3x+\cos (2x+\frac{\pi}{6})(\sin x+\sin 2x)=0$

$\Leftrightarrow \cos (2x+\frac{\pi}{6})(\sin x+\sin 2x +\sin 3x)=0$

$\Leftrightarrow \cos (2x+\frac{\pi}{6})(2\sin 2x.\sin x+\sin 2x )=0$

$\Leftrightarrow \sin 2x.\cos (2x+\frac{\pi}{6}).(2\sin x+1 )=0$

$\Leftrightarrow ............................$

Bài viết đã được chỉnh sửa nội dung bởi hoangtrong2305: 22-06-2012 - 23:06