Question

решить уравнение: 3cos 2x= 7sinx

Answer 1

$3cos2x = 7sinx \ \ 3(1 - 2sin^2x) = 7sinx \ \ 3 - 6sin^2x = 7sinx \ \ -6sin^2x - 7sinx + 3 = 0 \ \ 6sin^2x + 7sinx - 3 = 0$

Пусть $t = sinx, t in [-1; 1]$

$6t^2 + 7t - 3 = 0 \ \ D = 49 + 4 cdot 3 cdot 6 = 49 + 72 = 121 = 11^2 \ \ t_1 = dfrac{-7 + 11}{12} = dfrac{1}{3} \ \ t_2 = dfrac{-7-11}{12} = - dfrac{3}{2} - ne ud.$

Обратная замена:

$sinx = dfrac{1}{3} \ \ boxed{x = (-1)^{n}arcsin dfrac{1}{3} + pi n, n in Z}$