Question

Решите уравнение: 4cos²x - 2sin²x - 5cos x - 4 = 0

Answer 1

${sin}^{2} x = 1 - {cos}^{2} x \ 4 {cos}^{2} x - 2(1 - {cos}^{2} x) - 5cosx - 4 = 0 \ 4 {cos}^{2} x - 2 + 2 {cos}^{2} x - 5cosx - 4 = 0 \ 6 {cos}^{2} x - 5cosx - 6 = 0 \ cosx = t \ (- 1 leqslant t leqslant 1 )\ 6 {t}^{2} - 5t - 6 = 0 \ d = {b}^{2} - 4ac = 25 - 4 times 6 times ( - 6) = 169 \ t1 = frac{5 + 13}{12} = frac{18}{12} = frac{3}{2} \ t2 = frac{5 - 13}{12} = frac{ - 8}{12} = - frac{2}{3} \ cosx = - frac{2}{3} \ x = + - arccos( - frac{2}{3} ) + 2pi n$