If in a triangle ABC sines of angles A and B satisfies the equation $4x^2-2\sqrt{6}x+1=0$ ,then$\cos (A-B)=$ 

$\begin{array}{l}4x^2-2\sqrt{6}x+1=0\\ x=\frac{-\left(2\sqrt{6}\right)\pm \sqrt{24-16}}{2\left(4\right)}\\ =\frac{2\sqrt{6}\pm 2\sqrt{2}}{2\left(4\right)}=\frac{\sqrt{6}\pm \sqrt{2}}{4}\end{array}$

$=\sqrt{2}\left(\frac{\sqrt{3}\pm 1}{2\sqrt{2}.\sqrt{2}}\right)$

$=\frac{\sqrt{3}\pm 1}{2\sqrt{\{2}}$

Let $\sin A=\frac{\sqrt{3}+1}{2\sqrt{2}}\sin B=\frac{\sqrt{3}-1}{2\sqrt{2}}$

$\begin{array}{l}\sin A=\sin 75;\sin B=\cos 15\\ A=75;B=15\\ \cos (A-B)=\cos (75-15)=\cos 60=\frac{1}{2}\end{array}$