Please help a solve a trigo sum!!

The quadratic eqn in x, 3X^2 - (4cosA)X + 2sinA = 0, has equal real roots.

Find the value of A, which is between 0 degrees and 360 degrees.

Appreciate it if clear workings can be shown!

a = 3
b = -4 * cos(A)
c = 2 * sin(A)

b^2 - 4ac = 0
16 * cos(A)^2 - 4 * 3 * 2 * sin(A) = 0
16 * cos(A)^2 - 24 * sin(A) = 0
2 * cos(A)^2 - 3 * sin(A) = 0
2 * (1 - sin(A)^2) - 3 * sin(A) = 0
-2 * sin(A)^2 - 3 * sin(A) + 2 = 0
2 * sin(A)^2 + 3 * sin(A) - 2 = 0
sin(A) = (-3 +/- sqrt(9 + 16)) / 4
sin(A) = (-3 +/- 5) / 4
sin(A) = -8/4 , 2/4
sin(A) = -2 , 1/2
sin(A) = 1/2
A = 30 , 150

For equal roots , b² = 4ac
16cos²A = 24sinA
2 (1 - sin²A) = 3 sinA
2 - 2sin²A = 3 sinA
2 sin²A + 3 sinA - 2 = 0
(2 sinA - 1)(sinA + 2) = 0
sinA = 1/2
A = 30° , 150°