Question

Помогите решить уравнение:
cos4x+3sin2x-2=0

Answer 1

$cos4x + 3sin2x - 2 = 0 \ 1 - 2sin^22x + 3sin2x - 2 = 0 \ -2sin^22x + 3sin2x - 1 = 0 \ 2sin^22x - 3sin2x + 1 = 0 \ t = sin2, |t| leq 1 \ 2t^2 - 3t + 1 = 0 \ D = 9 - 2 cdot 4 = 1 \ \ t_1 = dfrac{3 + 1}{4} = 1 \ \ t_2 = dfrac{3 - 1}{4} = dfrac{1}{2}$
Обратная замена:
$sin2x = 1 \ \ 2x = dfrac{ pi }{2} + 2 pi n , n in Z \ \ boxed{x = dfrac{ pi }{4} + pi n , n in Z} \ \ sin2x = dfrac{1}{2} \ \ 2x = (-1)^{n} dfrac{ pi }{6} + pi k, k in Z \ \ boxed{x = (-1)^{n} dfrac{ pi }{12} + dfrac{ pi k}{2}, k in Z}$