Có $cos(\frac{5\pi}{12}-x)=cos(\frac{6\pi}{12}-\frac{\pi}{12}-x)=sin(x+\frac{\pi}{12})$
$PT \Leftrightarrow \sqrt2 . 2sinx.sin(x+\frac{\pi}{12})=1$
$\Leftrightarrow -\sqrt2 .[cos (2x+\frac{\pi}{12})-cos\frac{\pi}{12}]=1$
$\Leftrightarrow -\sqrt2.cos(2x+\pi/12)=1-\sqrt2.cos\frac{\pi}{12}=\frac{1-\sqrt3}2$
$\Leftrightarrow cos(2x+\pi/12)=\frac{\sqrt6-\sqrt2}4=cos(\pm \frac{5\pi}{12}) $
$x=\frac{\pi}6$ hoặc $x=-\frac{\pi}4$