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

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

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

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