Question

5-2sin(x)*cos(x)-5(sin(x)-cos(x)=0

Answer 1

$5-2\, sinx\cdot cosx-5(sinx-cosx)=0\\\\Zamena:\; \; t=sinx-cosx\; ,\\\\t^2=(sinx-cosx)^2=sin^2x+cos^2x-2\, sinx\cdot cosx=1-2\. sinx\cdot cosx\; \to \\\\2\, sinx\cdot cosx=1-t^2\\\\5-(1-t^2)-5t=0\\\\5-1+t^2-5t=0\\\\t^2-5t+4=0\; \; ,\; \; t_1=1\; ,\; t_2=4\; \; (teorema\; Vieta)\\\\a)\; \; sinx-cosx=1\; \; \to \; \; sinx-sin(\frac{\pi }{2}-x)=1\; ,\\\\2sin\frac{x-(\frac{\pi}{2}-x)}{2}\cdot cos\frac{x+(\frac{\pi}{2}-x)}{2}=1\\\\2sin(x-\frac{\pi}{4})\cdot cos\frac{\pi}{4}=1\\\\2sin(x-\frac{\pi }{4})\cdot \frac{\sqrt2}{2}=1$

$sin(x-\frac{\pi}{4})=\frac{\sqrt2}{2}\\\\x-\frac{\pi}{4}=(-1)^{n}\cdot \frac{\pi}{4}+\pi n\; ,\; n\in Z\\\\x=\frac{\pi }{4}+(-1)^{n}\cdot \frac{\pi }{4}+\pi n=\left [ {{\pi n\; ,\; esli\; n=2k-1\; ,\; k\in Z} \atop {\frac{\pi}{2}+\pi n\; ,\; esli\; n=2k\; ,\; k\in Z}} \right. \\\\b)\; \; sinx-cosx=4\\\\2sin(x-\frac{\pi }{4})\cdot \frac{\sqrt2}{2}=4\\\\sin(x-\frac{\pi}{4})=2\sqrt2>1\; \; \to \; \; net\; kornej\; ,\; t.k.\; |sin(x-\frac{\pi}{4})|\leq 1\\\\Otvet:\; \; x=\frac{\pi}{4}+(-1)^{n}\cdot\frac{\pi }{4}+\pi n\; ,\; n\in Z$

$\star \; \; cosx=sin(\frac{\pi}{2}-x)\; \; \star \\\\\star sinx-siny=2sin\frac{x-y}{2}\cdot cos\frac{x+y}{2}\; \; \star$