Решите уравнение: 2sin^2 2x-5cos2x=cos^2 2x

2Sin²2x - 5Cos2x = Cos²2x
2(1 - Cos²2x) - 5Cos2x - Cos²2x = 0
2 - 2Cos²2x - 5Cos2x - Cos²2x = 0
- 3Cos²2x - 5Cos2x + 2 = 0
3Cos²2x + 5Cos2x - 2 = 0
Сделаем замену Cos2x = m , - 1 ≤ m ≤ 1
3m² + 5m - 2 = 0
D = 5² - 4 * 3 * (- 2) = 25 + 24 = 49
$m _{1} = \frac{-5+ \sqrt{49} }{6}= \frac{-5+7}{6} = \frac{1}{3}$
$m _{2} = \frac{-5- \sqrt{49} }{6}= \frac{-5-7}{6} = \frac{-12}{6}= - 2$ -посторонний корень
Cos2x = $\frac{1}{3}$
2x = + - arcCos $\frac{1}{3}$ +2πn , n ∈ z
x = + - $\frac{1}{2} arcCos \frac{1}{3} * \pi n,$ n ∈ z