Помогите решить!! 3sinx-2sinx*cosx-3cosx=2

$3\sin x-2\sin x\cos x-3\cos x= 2\\ 3\sin x - 2\sin x\cos x - 3\cos x = 2\sin^2x +2\cos^2x\\ 3(\sin x-\cos x)-2\sin x\cos x-2(\sin^2x+\cos^2x)=0\\ 3(\sin x-\cos x)-\sin2x-2(\sin^2x+\cos^2x+\sin2x-\sin2x)=0\\ -2(\sin x-\cos x)^2-3(\sin x-\cos x)-3\sin 2x=0$

Пусть $\sin x - \cos x = t(|t|<=1)\\ 1-\sin2x=t^2\,\,\,\,=\ \textgreater \ \sin2x=1-t^2$
Получаем
$-2t^2+3t-3(1-t^2)=0\\ t^2+3t-3=0\\ D=9+12=21\\ t_1= \frac{-9+ \sqrt{21} }{2}\\ t_2=\frac{-9- \sqrt{21} }{2}$

Возвращаемся к замене
$\sin x-\cos x = \frac{-9+ \sqrt{21} }{2}\\ \sqrt{2} \sin (x- \frac{\pi}{4})=\frac{-9+ \sqrt{21} }{2}\\ x=(-1)^{k}\cdot \arcsin(\frac{-9+ \sqrt{21} }{2\sqrt{2}}) + \frac{\pi}{4} +\pi k,k \in Z$