Question

решите способом введения доп. аргумента уравнение:

sin2x+cos2x+1=0

Корень из 2 sinx=2-корень из 2cosx

Answer 1

$1)\; \; sin2x+cos2x+1=0\\\\sin2x+cos2x=-1\; |:\sqrt2\\\\\frac{1}{\sqrt2}\cdot sin2x+\frac{1}{\sqrt2}\cdot cos2x=-\frac{1}{\sqrt2}\\\\\star \; \; \frac{1}{\sqrt2}=\frac{\sqrt2}{2}=sin\frac{\pi}{4}=cos\frac{\pi}{4}\; \; \star \\\\cos\frac{\pi}{4}\cdot sin2x+sin\frac{\pi}{4}\cdot cos2x=-\frac{\sqrt2}{2}\\\\sin(2x+\frac{\pi}{4})=-\frac{\sqrt2}{2}\\\\2x+\frac{\pi}{4}=(-1)^{n}\cdot (-\frac{\pi}{4})+\pi n\; ,\; n\in Z\\\\2x=-\frac{\pi}{4}+(-1)^{n+1}\cdot \frac{\pi}{4}+\pi n=\frac{\pi}{4}\cdot ((-1)^{n+1}-1)+\pi n\; ,\; n\in Z\\\\x=\frac{\pi}{8}\cdot ((-1)^{n+1}-1)+\frac{\pi n}{2}\; ,\; n\in Z\\\\Otvet:\; \; x=\frac{\pi}{8}\cdot ((-1)^{n+1}-1)+\frac{\pi n}{2}\; ,\; n\in Z\; .$

$2)\; \; \sqrt2sinx=2-\sqrt2cosx\\\\\sqrt2sinx+\sqrt2cosx=2\, |:2\\\\\frac{\sqrt2}{2}sinx+\frac{\sqrt2}{2}cosx=1\\\\\star \; \; \frac{\sqrt2}{2}=sin\frac{\pi}{4}=cos\frac{\pi}{4}\; \; \star \\\\cos\frac{\pi}{4}\cdot sinx+sin\frac{\pi}{4}\cdot cosx=1\\\\sin(x+\frac{\pi}{4})=1\\\\x+\frac{\pi}{4}=\frac{\pi}{2}+2\pi n\; ,\; n\in Z\\\\x=\frac{\pi}{4}+2\pi n\; ,\; n\in Z\\\\Otvet:\; \; x=\frac{\pi}{4}+2\pi n\; ,\; n\in Z\; .$