Решите уравнение:

Решите задачу:

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

$b)\; \; sinx+cosx=-2\; |:\sqrt2\\\\\frac{1}{\sqrt2}=\frac{\sqrt2}{2}\\\\\frac{\sqrt2}{2}sinx+ \frac{\sqrt2}{2}cosx=-\frac{2}{\sqrt2}$

$cos \frac{\pi}{4}\cdot sinx+sin\frac{\pi}{4}\cdot cosx=-\sqrt2\; ,\; \; -\sqrt2\approx -1,41\\\\sin(x+\frac{\pi}{4})=-\sqrt2 \ \textless \ -1\; \; \; \; net\; reshenij,\; t.k.\; -1 \leq sin \alpha \leq 1\\\\Otvet:\; \; x=-\frac{\pi}{4}+\pi n,\; n\in Z\; .$