Question

найти корни уравнения

cos^2 x+3sinx-3=0/yf отрезке [-2pi;4pi]

Answer 1

$cos^2x + 3sinx - 3 = 0 \ \ 1 - sin^2x + 3sinx - 3 = 0 \ \ -sin^2x + 3sinx - 2 = 0 \ \ sin^2x - 3sinx + 2 = 0$
Пусть $t = sinx, t in [-1; 1].$
$t^2 - 3t + 2 = 0 \ \ t_1 + t_2 = 3 \ t_1 cdot t_2 = 2 \ \ t_1 = 2 - ne ud. \ t_2 = 1$
Обратная замена:
$sinx = 1 \ \ x = dfrac{ pi }{2} + 2 pi n, n in Z \ \ -2 pi leq dfrac{ pi }{2} + 2 pi n leq 4 pi , n in Z \ \ -4 pi leq pi + 4 pi n leq 8 pi , n in Z \ \ -4 leq 1 + 4n leq 8, n in Z \ \ -5 leq 4n leq 7, n in Z \ \ n = -1; 0; 1. \ \ x= - dfrac{3 pi }{2}; dfrac{ pi }{2} ; dfrac{5 pi }{2} .$