Question

6sin^2x - 5sinx-4=0 [-7 пи /2 ; -3 пи/2]

Answer 1

$6sin^2x - 5sinx - 4 = 0 \ \ t = sinx, t in [-1; 1] \ \ 6t^2 - 5t - 4 = 0 \ \ D = 25 + 4 cdot 4 cdot 6 = 96 + 25 = 121 = 11^2 \ \ t_1 = dfrac{5 + 11}{12} = dfrac{16}{12} - an extraneous root \ \ t_2 = dfrac{5 - 11}{12} = -dfrac{1}{2} \ \ Reverse substitution: \ \ sinx = -dfrac{1}{2} \ \ x = (-1)^{n+1} dfrac{pi} {6} + pi n, n in Z \ \ - dfrac{7 pi }{2} leq (-1)^{n+1} dfrac{pi} {6} + pi n leq - dfrac{3 pi }{2}, n in Z$
$-21 pi leq (-1)^{n + 1} pi + 6 pi n leq -9 pi , n in Z \ \ -21 leq (-1)^{n + 1} + 6n leq -9 , n in Z \ \ n = -3; -2 \ \ x_1 = (-1)^{-3 + 1} dfrac{pi }{6} - 3 pi = (-1)^{-2} dfrac{ pi} {6} - 3 pi = dfrac{pi }{6} - 3 pi = - dfrac{17 pi }{6} \ \ x_2 = (-1)^{-2 + 1}dfrac{pi }{6} - 2 pi = - dfrac{pi }{6} - 2 pi = - dfrac{13 pi }{6} \ \ boxed {Answer: x = - dfrac{17 pi }{6}; - dfrac{13 pi }{6}.}$