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:
$sin2x = 1 \\ \\ 2x = \dfrac{ \pi }{2} + 2 \pi n , \ n \in Z \\ \\ \boxed{x = \dfrac{ \pi }{4} + \pi n , \ n \in Z} \\ \\ sin2x = \dfrac{1}{2} \\ \\ 2x = (-1)^{n} \dfrac{ \pi }{6} + \pi k, \ k \in Z \\ \\ \boxed{x = (-1)^{n} \dfrac{ \pi }{12} + \ \dfrac{ \pi k}{2}, \ k \in Z}$