Помогите решить пожалуйста!! sin2x+5(cosx-sinx+1)=0

Решите задачу:

$sin2x+5(cosx-sinx+1)=0\\\\t=cosx-sinx\; \to \; \; t^2=cos^2x-2sinx\cdot cosx+sin^2x=1-sin2x\\\\\to \; \; sin2x=1-t^2\\\\1-t^2+5(t+1)=0\\\\t^2-5t-6=0\; \; \to \; \; t_1=-1\; ,\; \; t_2=6\\\\1)\; \; cosx-sinx=-1\; |:\sqrt2\\\\\frac{1}{\sqrt2}cosx-\frac{1}{\sqrt2}sinx=-\frac{1}{\sqrt2}\\\\sin\frac{\pi}{4}\cdot cosx-cos\frac{\pi}{4}\cdot sinx=-\frac{1}{\sqrt2}\\\\sin(\frac{\pi}{4}-x)=-\frac{1}{\sqrt2} \; \; \to \; \; sin(x-\frac{\pi}{4})=\frac{1}{\sqrt2}$

$x-\frac{\pi}{4}=(-1)^{n}arcsin\frac{1}{\sqrt2}+\pi n=(-1)^{n}\frac{\pi}{4}+\pi n,\; n\in Z\\\\x=\frac{\pi}{4}\cdot (1+(-1)^{n})+\pi n,\; n\in Z\\\\2)\; \; cosx-sinx=6\\\\\frac{1}{\sqrt2}cosx-\frac{1}{\sqrt2}sinx=\frac{6}{\sqrt2}\\\\sin(\frac{\pi}{4}-x)=\frac{6}{\sqrt2}\\\\sin(\frac{\pi}{4}-x)=3\sqrt2\ \textgreater \ 1\; \; \Rightarrow \; \; net\; reshenij$