Решить тригонометрическое уравнение sin(x) - cos(x)2x = 2

[latex]\sin x - \cos2x = 2 \\ \sin x - (\cos^2 x -\sin ^2 x) = 2 \\ \sin x - (\cos^2 x +\sin^2x-2\sin ^2 x) = 2 \\ \sin x - (1-2\sin ^2 x) = 2 \\ \sin x - 1+2\sin ^2 x = 2 \\ 2\sin^2x+\sin x-3=0\\ 2\sin^2x-2\sin x+3\sin x-3=0\\ 2\sin x(\sin x-1)+3(\sin x-1)=0\\ (2\sin x+3)(\sin x-1)=0\\\\ 2\sin x+3=0\\ 2\sin x=-3\\ \sin x =-\dfrac{3}{2}\\ x\in\emptyset\\\\ \sin x-1=0\\ \sin x=1\\ \boxed{x=\dfrac{\pi}{2}+2k\pi, k\in\mathbb{Z}} [/latex]