|
|
Ta có
$\tan(x+\dfrac{\pi}{3})=\cot(\dfrac{\pi}{2}-(x+\dfrac{\pi}{3}))=\cot(\dfrac{\pi}{6}-x)=-\cot(x-\dfrac{\pi}{6})$
$\Rightarrow \tan(x-\dfrac{\pi}{6})\tan(x+\dfrac{\pi}{3})=-\tan(x-\dfrac{\pi}{6})\cot(x-\dfrac{\pi}{6})=-1$
PT
$\Leftrightarrow -\sin3x =\sin x +\sin2x$
$\Leftrightarrow\sin x+\sin3x+\sin2x=0$
$\Leftrightarrow 2\sin 2x\cos x+\sin2x=0$
$\Leftrightarrow \sin 2x(2\cos x+1)=0$
$\Leftrightarrow \left[ {\begin{matrix} \sin 2x =0\\ \cos x =-1/2 \end{matrix}} \right.$
|