Решите уравнение1-sin2x= -(sinx+cosx)найдите все корни на промежутке -3П/2;П

[latex]1-sin2x=-(sinx+cosx)\\\\t=sinx+cosx\\\\t^2=(sin^2x+cos^2x)+2sinxcosx=1+sin2x\; \; \to \; \; sin2x=t^2-1\\\\\\1-(t^2-1)=-t\\\\t^2-t-2=0\\\\t_1=2,t_2=-1\; (teor.\; Vieta)\\\\1)\; sinx+cosx=2|:\sqrt2[/latex] [latex]\frac{1}{\sqrt2}sinx+\frac{1}{\sqrt2}cosx=\frac{2}{\sqrt2}\\\\cos\frac{\pi}{4}\cdot sinx+sin\frac{\pi}{4}\cdot cosx=\frac{2}{\sqrt2}\\\\sin(x+\frac{\pi}{4})=\frac{2}{\sqrt2}>1\; \to \; net\; reshenij\\\\2)\; sinx+cosx=-1|:\sqrt2\\\\\frac{1}{\sqrt2}sinx+\frac{1}{\sqrt2}cosx=-\frac{1}{\sqrt2}\\\\sin(x+\frac{\pi}{4})=-\frac{1}{\sqrt2}\\\\x+\frac{\pi}{4}=(-1)^{n+1}\cdot \frac{\pi}{4}+\pi n,\; n\in Z\\\\x=(-1)^{n+1}\cdot \frac{\pi}{4}-\frac{\pi}{4}+\pi n [/latex] [latex]x= \left \{ {{\pi n,esli\; n-nechetnoe} \atop {-\frac{\pi}{2}+\pi n,esli\; n- chetnoe}} \right. [/latex]