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

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

Bull·shit!!!

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

(1)x²+x+1 [tex]\ge[/tex] -1 => x²+x+2 [tex]\ge[/tex] 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 [tex]\le[/tex] 1 => x²+x [tex]\le[/tex] 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

Any suggestions?

Prove that the equation
[tex]sinx - x^2 - x - 1 = 0[/tex]
has no real solutions.