Решить уравнение sin х + cos х = 1 − sin 2х. А) (−1)^n*pi/4+ 2; В) 2pi*n/3+ 2pi*n; Б)...

$sinx+cosx=1-sin2x\\\\Zamena:\; \; t=sinx+cosx\; \to \; \; t^2=(sinx+cosx)^2\\\\t^2=sin^2x+cos^2x+2sinx\cdot cosx=1+sin2x\; \to \; \; sin2x=t^2-1\\\\t=1-(t^2-1)\; ,\; \; \; t=1-t^2+1\\\\t^2+t-2=0\\\\t_1=-2\; ,\; \; t_2=1\; \; (teorema\; Vieta)\\\\a)\; \; sinx+cosx=-2\; |:\sqrt2\\\\\frac{1}{\sqrt2}\cdot sinx+ \frac{1}{\sqrt2}\cdot cosx=- \frac{2}{\sqrt2}\\\\cos\frac{\pi}{4}\cdot sinx+sin \frac{\pi}{4}\cdot cosx=-\sqrt2\\\\sin(x+\frac{\pi}{4})=-\sqrt2\\\\-\sqrt2\approx -1,4<-1\; \; ,\; \; a\; \; -1\leq sin(x+\frac{\pi}{4})\leq 1\; \; \Rightarrow$

$sin(x+\frac{\pi}{4})=-\sqrt2\; \; ne\; imeet\; reshenij\; ,\; x\in \varnothing \\\\b)\; \; sinx+cosx=1\; |:\sqrt2\\\\\frac{1}{\sqrt2}\cdot sinx+ \frac{1}{\sqrt2}\cdot cosx= \frac{1}{\sqrt2}\\\\sin(x+\frac{\pi}{4})=\frac{1}{\sqrt2}\; ,\; \; \; \; \frac{1}{\sqrt2}=\frac{\sqrt2}{\sqrt2\cdot \sqrt2}=\frac{\sqrt2}{2}\\\\sin(x+\frac{\pi}{4})=\frac{\sqrt2}{2}\\\\x+\frac{\pi}{4}=(-1)^{n}\cdot arcsin\frac{\sqrt2}{2}+\pi n=(-1)^{n}\cdot \frac{\pi }{4}+\pi n\; , \; n\in Z\\\\x=-\frac{\pi}{4}+(-1)^{n}\cdot (-1){n}\cdot \frac{\pi}{4}+\pi n$

$\underline {x=((-1)^{n}-1)\cdot \frac{\pi}{4}+\pi n,\; n\in Z}$

Ответ: Г) .