3 sin x = 2 ( 1 - sin² x )
3 sin x = 2 - 2 sin² x
2 sin² x + 3 sin x - 2 = 0
Substitution: t = sin x
2 t² + 3 t - 2 = 0
t 1/2 = \(\frac{-3+/- \sqrt{9+16} }{4} = \frac{-3+/-5}{4}\)
t 1 = 1/2
t 2 = - 2 ( this solution is not acceptable )
sin x = 1/2
Answer:
x 1 = π / 6 + 2 kπ, x 2 = 5 π / 6 + 2 k π, k ∈ Z
