In a triangle $ABC$, if $a^4+b^4+c^4=2a^2{b}^2+b^2{c}^2+2c^2{a}^2$, then $\sin A=$ 

Solution:

$a^4+b^4+c^4-2a^2\left(b^2+c^2\right)=b^2{c}^2$

$\begin{array}{l}\Rightarrow {\left(b^2+c^2-a^2\right)}^2=3b^2{c}^2\\ \Rightarrow b^2+c^2-a^2=\pm \sqrt{3}bc\Rightarrow 2bc\cos A=\pm \sqrt{3}bc\\ \Rightarrow \cos A=\frac{\pm \sqrt{3}}{2}\Rightarrow {\sin }^{2}A=1-\frac{3}{4}=\frac{1}{4}\Rightarrow \sin A=\frac{1}{2}\end{array}$