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

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

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