Here's my rather (long) take on this one......
So we have
a(1-x^2) + 2bx + (1 + x^2)c = 0 and 3c = a + 3b → a = 3c - 3b → a = 3(c-b)
So
3(c - b)(1- x^2) + 2bx + (1 +x^2)c = 0 factor the negative from(b-c) and applying it to (1 - x^2) and we have
3(b-c)(x^2 -1) + 2bx + (1+x^2)c = 0 rearranging and factoring we have
[(3(b-c) + c]x^2 + 2bx + [c - 3(b-c)] = 0
(3b -2c)x^2 + 2bx + (4c - 3b) = 0
Now, because this quadratic has a repeated solution, it must mean that its discriminant = 0
So, we have
4b^2 - 4(3b - 2c) (4c - 3b) = 0 divide through by 4
b^2 - (3b - 2c)(4c - 3b) = 0 factor the negative from (3b - 2c)
b^2 + (2c - 3b)(4c - 3b) = 0
b^2 + 8c^2 - 18bc + 9b^2 = 0 combine like terms
10b^2 -18bc + 8c^2 = 0 divide through by 2
5b^2 - 9bc + 4c^2 = 0 factor
(5b - 4c) ( b - c) = 0
So, either.......
b = c ..... but this is impossible because it makes a = 0 (remember that a = 3(c-b) ....)
or
(5b - 4c) = 0 ....... in which case ...... [ 5b = 4c ] → [ b = (4/5) c ]
So a = 3[c - (4/5)c] = [3(1/5)c] = (3/5)c
Now, let c be some convenient length, say, 5 ...... this means that b = 4 and a = 3
And, as Melody has correctly said, we have a classic 3-4-5 "Pythagorean Triple" right triangle !!!
And we could find one angle using the Law of Cosines.....the other angle would then be easily found as well as the sines of A and B .......(This part of the problem isn't particularly challenging !!!)
Note the interesting thing here... the value of the single "solution" to the quadratic isn't of any import....the fact that the polynomial only has one solution allows us to work with its discriminant to obtain the "answer" to an entirely different problem !!!
That's my story and I'm sticking to it............
