Кто может помочь с С1 ? только "а" часть)

$\sqrt{sin2x}=\sqrt{cosx-sinx-1},\; \; \; OOF:\; \left \{ {{sin2x \geq 0} \atop {cosx-sinx-1 \geq 0}} \right. \\\\sin2x=cosx-sinx-1\\\\1+2sinx\cdot cosx=cosx-sinx\\\\Vurazim:\\\\(cosx-sinx)^2=cos^2x+sin^2x-2sinx\cdot cosx=1-2sinx\cdot cosx\; \to \\\\2sinx\cdot cosx=1-(cosx-sinx)^2\\\\Podstavim\; v\; yravnenie:\; 1+(1-(cosx-sinx)^2=cosx-sinx\\\\t=cosx-sinx,\; t^2+t-2=0,\\\\t_1=-2,\; t_2=1$

0\; \to cosx-sinx>1\\\\2)cosx-sinx=1|:\sqrt2\\\\\frac{1}{\sqrt2}cosx-\frac{1}{\sqrt2}sinx=\frac{1}{\sqrt2}\\\\sin\frac{\pi}{4}cosx-cos\frac{\pi}{4}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+1}\frac{\pi}{4}+\pi n,\; n\in Z\\\\x=\frac{\pi}{4}+(-1)^{n+1}\frac{\pi}{4}+\pi n=\frac{\pi}{4}((-1)^{n+1}+1)+\pi n" alt="1)cosx-sinx=-2\; net\; reshenij,\\\\tak\; kak\; iz\; OOF:cosx-sinx-1>0\; \to cosx-sinx>1\\\\2)cosx-sinx=1|:\sqrt2\\\\\frac{1}{\sqrt2}cosx-\frac{1}{\sqrt2}sinx=\frac{1}{\sqrt2}\\\\sin\frac{\pi}{4}cosx-cos\frac{\pi}{4}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+1}\frac{\pi}{4}+\pi n,\; n\in Z\\\\x=\frac{\pi}{4}+(-1)^{n+1}\frac{\pi}{4}+\pi n=\frac{\pi}{4}((-1)^{n+1}+1)+\pi n" align="absmiddle" class="latex-formula">