Question

2sinxcosx+sinx+cosx+1=0

Answer 1

2sinx*cos+sinx+cosx+1=0
(2sinx*cosx+1)+(sinx+cosx)=0
(2sinx*cosx+1)+(√(sinx+cosx))²=0
(2sinx*cosx+1)+√(sin²x+2sinx*cosx+cos²x)=0
(2sinx*cosx+1)+√(2sinx*cosx+1)=0
замена переменной:
√(2sinx*cosx+1)=t, 
t²+t=0
t₁=0, t₂=-1
обратная замена:
1. t=0, 2sinx*cosx+1=0, sin2x=-1
2x=-π/2+2πn, n∈Z |:2
x₁=-π/4+πn. n∈Z
2. t=-1, √(2sinx*cosx+1)=-1. 2sinx*cosx+1=1, 2sinx*cosx=0
sinx=0 или cosx=0
x₂=πn, n∈Z
x₃=π/2+πn, n∈Z

Answer 2

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

$sin(x+\frac{\pi}{4})=-\frac{\sqrt2}{2}$

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