Question

Решите пожалуйста:

2sinx^2-3cosx-3=0​

Answer 1

2Sin²x - 3Cosx - 3 = 0

2(1 - Cos²x) - 3Cosx - 3 = 0

2 - 2Cos²x - 3Cosx - 3 = 0

- 2Cos²x - 3Cosx - 1 = 0

2Cos²x + 3Cosx + 1 = 0

Сделаем замену : Cosx = m, - 1 ≤ m ≤ 1

2m² + 3m + 1 = 0

D = 3² - 4 * 2 * 1 = 9 - 8 = 1

$m_{1}=frac{-3+1}{4}=-frac{1}{2}\\m_{2}=frac{-3-1}{4}=-1\\1)Cosx=-frac{1}{2}\\x=pm arcCos(-frac{1}{2})+2pi n,nin z\\x=pm (pi - arcCosfrac{1}{2})+2pi n,nin z\\x=pm (pi -frac{pi }{3})+2pi n,nin z\\x=pm frac{2pi }{3}+2pi n,nin z$

$2)Cosx=-1\\x=pi +2pi n,nin z$