# Prove that the equation has no real solutions

### Prove that the equation has no real solutions

Prove that the equation
$sinx - x^2 - x - 1 = 0$
has no real solutions.

Math Tutor

Posts: 313
Joined: Sun Oct 09, 2005 11:37 am

### Re: Prove that the equation has no real solutions

Any suggestions?
Math Tutor

Posts: 313
Joined: Sun Oct 09, 2005 11:37 am

### Re: Prove that the equation has no real solutions

sinx=x²+x+1 but -1 $\le$ sinx $\le$ 1
so we have the system
lx²+x+1 $\ge$ -1 (1)
lx²+x+1 $\le$ 1 (2)

(1)x²+x+1 $\ge$ -1 => x²+x+2 $\ge$ 0
Let's suppose that it isi equality then it doesn't have solutions.If it is only x²+x+2>0 the D<0 and a>0 so every x is possible
(2)x²+x+1 $\le$ 1 => x²+x $\le$ 0 .Lets suppose it is an equality then the roots are -1,0 if we put them into sinx=x²+x+1 we will see that it has no solutins.Now if x²+x<0 well it more than obvious that it is impossible.With that the system has no real solutions and the whole problem
idontknow

Posts: 2
Joined: Sat Jun 30, 2012 6:08 am

### Re: Prove that the equation has no real solutions

idontknow wrote:(2)x2+x+1 < 1 => x2+x < 0 .Lets suppose it is an equality then the roots are -1,0 if we put them into sinx=x2+x+1 we will see that it has no solutins.Now if x2+x<0 well it more than obvious that it is impossible.With that the system has no real solutions and the whole problem

Bull·shit!!!
shemet

Posts: 6
Joined: Tue Jun 14, 2011 7:59 am

### Re: Prove that the equation has no real solutions

Just take a look at the graph of f(x) = x^2+x+1 and g(x) = sin (x) and you're almost done. In the critical area
when x in [-1,0], the values of g(x) = sin (x) are negative. For the rest of the values there is no worry since
max(sin(x)) = 1.

YS
Guest