Respuesta :

cos(2x) = 1 - 2sin^2x

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

sin(x) = {1/2, 1} 
In the range [0, 2π], x = {π/6, π/2, 5π/6}