Question

6sin квадрат x + 5 cos x - 7=0

Answer 1

$6sin^2x + 5cosx - 7 = 0 \ 6sin^2x - 6 + 5cosx - 1 = 0 \ -6cos^2x + 5cosx - 1 = 0 \ 6cos^2x - 5cosx + 1 = 0$
Пусть $t = cosx, t in [-1; 1].$
$6t^2 - 5t + 1 = 0 \ D = 25 - 4 cdot 6 = 1 \ \ t_1 = dfrac{5 + 1}{12} = dfrac{1}{2} \ \ t_2 = dfrac{5-1}{12} = dfrac{1}{3}$
Обратная замена:
$cosx = dfrac{1}{2} \ \ boxed{x= pm frac{ pi }{3} +2 pi n, n in Z} \ \ cosx = dfrac{1}{3} \ \ boxed{x = pm arccos dfrac{1}{3} + 2 pi n, n in Z.}$

Answer 2

6sin²x + 5cosx - 7 = 0 ⇔ 6cos²x - 5cosx + 1 =0 ⇔ [ cosx =1/3 ; cosx =1/2 .