In a triangle ABC, a, b and c are the opposite sides of each angles ,and 2a*SinA=(2b+c)*SinB+(2c+b)*SinC
(I） try to the angle A
$$\\2aSinA=(2b+c)SinB+(2c+b)SinC\\
0=(2b+c)SinB+(2c+b)SinC-2aSinA\\$$
Using the sine rule I know that
$$\\SinB=\frac{bSinA}{a}\qquadand\qquad SinC=\frac{cSinA}{a}\\\\
$Substituting I get$\\\\$$
$$\\0=(2b+C)\frac{bSinA}{a}+(2c+b)\frac{cSinA}{a}-2aSinA\\\\
0=SinA[(2b+C)\frac{b}{a}+(2c+b)\frac{c}{a}-2a]\\\\
0=\frac{SinA}{a}[(2b+c)b+(2c+b)c-2a^2]\\\\
$Neither sinA nor A can be 0 so$\\\\$$
$$\\0=[2b^2+cb+2c^2+bc-2a^2]\\\\
0=[2b^2+2c^2-2a^2+2bc]\\\\
0=b^2+c^2-a^2+bc\\\\
-bc=b^2+c^2-a^2\\\\
-\frac{1}{2}=\frac{b^2+c^2-a^2}{2bc}\\\\
-\frac{1}{2}=cosA\qquad\qquad \mbox{Using cosine rule}\\\\
A=\pi-\frac{\pi}{3}\\\\
A=\frac{2\pi}{3}\\\\$$
.
+118734 
