Question

Даю 20 баллов. Решите тригонометрическое уравнение, желательно с пояснением.
Sin2x+√2*Sin(x-pi/4)=1

Answer 1

$sin2x+\sqrt2\, sin(x-\frac{\pi}{4})=1\\\\sin2x+\sqrt2\cdot (sinx\cdot cos\frac{\pi}{4}-cosx\cdot sin\frac{\pi}{4})=1\\\\sin2x+\sqrt2\cdot (\frac{\sqrt2}{2}\cdot sinx-\frac{\sqrt2}{2}\cdot cosx)=1\\\\2\, sinx\cdot cosx+sinx-cosx=sin^2x+cos^2x\\\\sinx-cosx=sin^2x-2sinx\cdot cosx+cos^2x\\\\sinx-cosx=(sinx-cosx)^2\\\\(sinx-cosx)^2-(sinx-cosx)=0\\\\(sinx-cosx)\cdot (sinx-cosx-1)=0\\\\a)\; \; sinx-cosx=0\, |:cosx\ne 0\\\\tgx=1\; ,\; \underline {x=\frac{\pi}{4}+\pi n,\; n\in Z}$

$b)\; \; sinx-cosx=1\, |:\sqrt2\\\\\frac{1}{\sqrt2}\cdot sinx-\frac{1}{\sqrt2}\cdot cosx=\frac{1}{\sqrt2}\\\\cos\frac{\pi}{4}\cdot sinx-sin\frac{\pi}{4}\cdot cosx=\frac{1}{\sqrt2}\; ,\; \; (\frac{1}{\sqrt2}=\frac{\sqrt2}{2})\\\\sin(x-\frac{\pi}{4})=\frac{1}{\sqrt2}\; ,\; \; x-\frac{\pi }{4}=(-1)^{n}\cdot \frac{\pi }{4}+\pi k\; ,\; k\in Z\\\\\underline {x=\frac{\pi}{4}\cdot (1+(-1)^{n})+\pi k,\; k\in Z}$