Đáp án: $ x=\dfrac{\pi}{3}+\dfrac{k\pi}{2}, x\notin \{\dfrac{\pi}{3}+k\pi, \dfrac{5\pi}{6}+k\pi, \}$
Giải thích các bước giải:
ĐKXĐ: $ x\notin \{\dfrac{\pi}{3}+k\pi, \dfrac{5\pi}{6}+k\pi, \}$
Ta có:
$\tan(x+\dfrac{\pi}{6})\cdot \tan(x-\dfrac{\pi}{3})=1$
$\to \dfrac{\sin(x+\dfrac{\pi}{6})}{\cos(x+\dfrac{\pi}{6})}\cdot \dfrac{\sin(x-\dfrac{\pi}{3})}{\cos(x-\dfrac{\pi}{3})}=1$
$\to \sin(x+\dfrac{\pi}{6})\cdot \sin(x-\dfrac{\pi}{3})=\cos(x+\dfrac{\pi}{6})\cdot \cos(x-\dfrac{\pi}{3})$
$\to \cos(x+\dfrac{\pi}{6})\cdot \cos(x-\dfrac{\pi}{3})-\sin(x+\dfrac{\pi}{6})\cdot \sin(x-\dfrac{\pi}{3})=0$
$\to \cos(x+\dfrac{\pi}{6}+x-\dfrac{\pi}{3})=0$
$\to \cos(2x-\dfrac{\pi}{6})=0$
$\to 2x-\dfrac{\pi}{6}=\dfrac{\pi}{2}+k\pi$
$\to x=\dfrac{\pi}{3}+\dfrac{k\pi}{2}$
