Question

va roggg sa se rezolve ecuatia : arcsin(x^2+5|x|-5,5)=pi/6

Answer 1

[tex]\arcsin(x^2+5|x|-5,5)=\dfrac{\pi}{6}|\sin()\\ x^2+5|x|-5,5=\dfrac{1}{2}|\cdot 2\\ 2x^2+10|x|-11=1\\ 2x^2+10|x|-12=0|:2\\ x^2+5|x|-6=0\\ \text{Pentru}\ x\geq0: |x|=x\\ x^2+5x-6=0\\ \Delta=25+24=49\Rightarrow \sqrt{\Delta}=7\\ x_1=\dfrac{-5+7}{2}=1 \\ x_2=\dfrac{-5-7}{2}=-6\ \textless \ 0(\text{nu convine})\\ \text{Pentru}\ x\ \textless \ 0: |x|=-x\\ x^2-5x-6=0\\ \Delta=25+24=49\Rightarrow \sqrt{\Delta}=7\\ x_3=\dfrac{5+7}{2}=6\ \textgreater \ 0(\text{nu convine})\\ x_4=\dfrac{5-7}{2}=-1\\ S:x\in \{\pm 1\}[/tex]

Answer 2

sin (arcsin(x²+5|x|-5,5))=sin π/6=1/2

x²+5|x|-5,5=1/2
2x²+10|x|-11=1

2x²+10|x|-11-1=0
2x²+10|x|-12=0

x²+5|x|-6=0


x²+5x-6 =0 si x≥0
x1,2=(-5+-√(25+24))/2

x1,2=(-5+-√49)/2
cu x>0 ramanedoar x1= (-5+7)/2=2/2=1

x²-5x-6=0 si x<0
x3,4= (5+-√49)/2
 cum x<0 ramane doar x3=(5-7)/2=-2/2=-1

deci S= {-1;1}
verificare ;
arcsin (1+5-5,5)=arcsin(0,5)=π/6 adevarat, problema este bine rezolvata