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Решите задачу:

$sinx+cosx=sin^2x-cos^2x\; ,\; \; -\frac{3\pi }{2} \leq x \leq 0\\\\sinx+cosx=(sinx-cosx)(sinx+cosx)\\\\(sinx+cosx)(1-sinx+cosx)=0\\\\1)\quad sinx+cosx=0|:cosx\ne 0\\\\tgx=-1,\; \; x=-\frac{\pi}{4}+\pi n,n\in Z\\\\2)\quad 1-sinx+cosx=0\\\\sinx-cosx=1|:\sqrt2\\\\\frac{1}{\sqrt2}sinx-\frac{1}{\sqrt2}cosx=\frac{1}{\sqrt2}$

$cos\frac{\pi}{4}sinx-sin\frac{\pi}{4}cosx=\frac{1}{\sqrt2}\\\\sin(x-\frac{\pi}{4})=\frac{1}{\sqrt2}\\\\x-\frac{\pi}{4}=(-1)^{k}\frac{\pi}{4}+\pi k,k\in Z\\\\x=\frac{\pi}{4}+(-1)^{k}\frac{\pi}{4}+\pi k=\frac{\pi}{4}(1+(-1)^k)+\pi k= \left \{ {{\pi k,\; esli\; k=2m} \atop {\frac{\pi}{2}+\pi k,\; esli\; k=2m-1}} \right. \\\\Esli\; -\frac{3\pi }{2} \leq x \leq 0,\; to\; x=-\frac{\pi}{4}\; i\; x=-\frac{5\pi }{4}, x=-\frac{\pi}{2}$