3 sin x = 1 + cos 2x
use the trig identity: cos(2x) = 1 - 2sin^2(x)
3sin(x) = 1 + 1 - 2sin^2(x) now treat this as a quadratic equation.
3u = 2 - 2u^2 here u = sin(x)
2u^2 + 3u - 2 = 0
(2u - 1)(u + 2) = 0
u = 1/2, u = -2
sin(x) = 1/2, sin(x) = -2 <--- ignore this last solution, since although it is a solution to the quadratic equation it cannot be a solution to the trig equation, because |sin(x)| <= 1.
sin(x) = 0.5, therefore x = pi/6, and x = 5pi/6