katmyloryleBeretos
katmyloryleBeretos katmyloryleBeretos
  • 02-09-2016
  • Mathematics
contestada

Solve 3sinx = 2cos^2x .

Respuesta :

HomertheGenius
HomertheGenius HomertheGenius
  • 02-09-2016
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 = [tex] \frac{-3+/- \sqrt{9+16} }{4} = \frac{-3+/-5}{4} [/tex]
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 
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