Question

Найдите количество корней уравнения(1рис) на промежутке (2рис)

Answer 1

$sin(3\pi -2x)+1=cos(\frac{\pi}{2}-x)-cos(\pi -x)\\\\sin2x+1=sinx+cosx\\\\zamena:\; \; t=sinx+cosx\; ,\\\\t^2=sin^2x+cos^2x+2sinx\cdot cosx=1+sin2x\; \; \Rightarrow \quad sin2x=t^2-1\\\\t^2-1+1=t\\\\t^2-t=0\; ,\; \; t\, (t-1)=0\; ,\; \; t_1=0\; ,\; \; t_2=1\\\\a)\; \; sinx+cosx=0\; |:cosx\ne 0\\\\tgx+1=0\; ,\; \; tgx=-1\\\\x=-\frac{\pi}{4}+\pi n,\; n\in Z\\\\b)\; \; sinx+cosx=1\\\\\sqrt2\cdot (\frac{1}{\sqrt2}\cdot sinx+ \frac{1}{\sqrt2}\cdot cosx )=1\\\\\sqrt2\cdot (cos\frac{\pi }{4}\cdot sinx+sin\frac{\pi}{4}\cdot cosx)=1\\\\sin(x+ \frac{\pi }{4})=\frac{1}{\sqrt2}$

$x+\frac{\pi }{4}=(-1)^{k}\cdot \frac{\pi }{4}+\pi k,\; k\in Z\\\\x=-\frac{\pi }{4}+(-1)^{k}\cdot \frac{\pi }{4}+\pi k,\; k\in Z\\\\x=((-1)^{k}-1)\cdot \frac{\pi }{4}+\pi k\; ,\; k\in Z \\\\Otvet:\; \; x=((-1)^{k}-1)\cdot \frac{\pi}{4}+\pi k\; ;\; x=-\frac{\pi}{4}+\pi n\; ;\; \; n,k\in Z\; .\\\\ili\; \; x=\left [ {{\pi k\; ,\; k-nechetnoe\; \; } \atop {-\frac{\pi}{2}+\pi k\; ,\; k-chetnoe}} \right. \; ;\; x=-\frac{\pi}{4}+\pi n\; ,\; n\in Z\; .$

$x\in [\, \frac{\pi}{2},2\pi ):\; \; \; x=\frac{\pi}{2}\; ,\; \frac{3\pi}{4}\; ,\; \pi \; ,\; \frac{3\pi}{2}\; ,\frac{7\pi}{4}\; .$