Respuesta :

nobillionaireNobley
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)

Otras preguntas