2sinxcosx+sinx+cosx+1=0

Решите задачу:

$2sinx\cdot cosx+sinx+cosx+1=0\\\\sin2x+sinx+cosx+1=0\\\\t=sinx+cosx\; ,\; \; t^2=sin^2x+cos^2x+2sinx\cdot cosx=1+sin2x\\\\sin2x=t^2-1\\\\t^2-1+t+1=0\\\\t\cdot (t+1)=0\\\\t_1=0\; ,\; \; t_2=-1\\\\a)\quad sinx+cosx=0\, |:cosx\ne 0\\\\tgx+1=0\\\\tgx=-1\\\\x=-\frac{\pi}{4}+\pi n,\; n\in Z\\\\b)\quad sinx+cosx=-1\, |:\sqrt2\\\\ \frac{1}{\sqrt2}\cdot sinx+ \frac{1}{\sqrt2}\cdot cosx =-\frac{1}{\sqrt2}\\\\cos\frac{\pi}{4}\cdot sinx+sin\frac{\pi}{4}\cdot cosx=-\frac{1}{\sqrt2}$

$sin(x+\frac{\pi}{4})=-\frac{\sqrt2}{2}$

$x+\frac{\pi}{4}=(-1)^{m}\cdot (-\frac{\pi}{4})+\pi m,\; m\in Z\\\\x+\frac{\pi}{4}=(-1)^{m+1}\cdot \frac{\pi}{4}+\pi m\; ,m\in Z\\\\x=-\frac{\pi}{4}+(-1)^{m+1}\cdot \frac{\pi}{4}+\pi m,\; m\in Z\\\\x=\frac{\pi}{4}\cdot ((-1)^{m+1}-1)+\pi m,\; m\in Z$