Question

а) Решите уравнение
$\sqrt{2}cos2x - 2cos(3\pi /2 + x) - \sqrt{2} = 0$

б) Укажите корни этого уравнения, принадлежащие отрезку $[3\pi/2;3\pi]$

Answer 1

$\sqrt{2} \cos2x - 2\cos ( \frac{3\pi}{2} + x) - \sqrt{2} = 0 \\ \sqrt{2} \cos2x - 2\sin x - \sqrt{2} = 0 \\ \cos 2x - \sqrt{2} \sin x - 1 = 0 \\ 1 - 2\sin^2 x - \sqrt{2} \sin x - 1 = 0 \\ 2\sin^2 x + \sqrt{2} \sin x = 0 \\ \sqrt{2} \sin^2 x + \sin x = 0 \\ \sin x ( \sqrt{2} \sin x + 1) = 0 \\ \sin x = 0 \\ x = \pi n , n \in Z \\ \sqrt{2} \sin x = -1 \\ \sin x = - \frac{ \sqrt{2} }{2} \\ x = (-1)^{n+1} \frac{\pi}{4} + \pi n, n \in Z.$