Find the Solution of 2cos^2x + 3sin x = 0

2cos^2 x + 3sin x = 0

2 (1 - sin^2 x) + 3sin x = 0

2 - 2sin^2 x + 3sin x = 0

2sin^2 x - 3sin x - 2 = 0

Let sin x = m, then

2m^2 - 3m - 2 = 0

2m^2 + m - 4m - 2 = 0

m (2m + 1) - 2 (2m + 1) = 0

(m - 2) (2m + 1) = 0

m = 2 or m = - 1/2

Now, sin x = - 1/2

Therefore, x = 1/6 (12nπ - π) and 1/6 (12nπ + 7π)