giorgio

I understant that trial values for x would give me " pi/2 " ,but I'm trying to find out another method. Thank for the answer.

Any suggest about how to solve this ecuation will be appretiated [math] x(\csc(x)+1)=\pi[/math] Thanks in advance.

Hi It is well knowm by the theorem of the reminder that: [math]D(x)=d(x)q(x)+r(x)[/math] In this example: .......................If [math](2x^{119}+1)/(x^2-x+1)[/math] then figure out the reminder. Therefore I can afirm that the reminder has the form of :[math]r(x)=ax+b[/math], because [math]d(x)>r(x)[/math] and [math]d(x)=x^2-x+1[/math] And the original problem could be written as: ..................[math]2x^{119}+1=(x^2-x+1)q(x)+ax+b[/math].........(I) To this point I need two solution of:[math]d(x)=0[/math] so I can find the values of a and b. According to what I think: ...............If [math](2x^{119}+1) = D(x)[/math] and [math]d(x)=(x^2-x+1)[/math] so in order to find the solution :[math] x^2-x+1=0[/math]......(II) .............................[math](x+1)(x^2-x+1)=0[/math] ....................................[math]x^3+1=0 [/math] ..............................................[math]x^3=-1 [/math]........(III) .................................. replacing (III) on [math]D(x)[/math]: ....................................[math]2(-1)x^2+1= -2x^2+1 [/math] .................................but in (II) we'll have: [math] -x^2=-x+1[/math] ......................................so [math]-2x^2+1=-2x+3[/math] ........................................finally [math]-2x+3=r(x) [/math] ........................................[math]ax+b=-2x+3[/math] I just found the above anwer substituting values for another, but really would like to know if I can find it in the form (I) below: ...................................[math]2x^{119}+1=(x^2-x+1)q(x)+ax+b.............(I)[/math] ........................................and if I try to fatorise the way I did: ..........................................[math]2x^{119}+1=(x^2-x+1)(x+1)(x+1)^{-1}q(x)+ax+b[/math] ...........................................[math]2x^{119}+1=(x^3+1)(x+1)^{-1}q(x)+ax+b[/math] .................................................When I change x=-1 ............................................... the expression: [math](x+1)^{-1}[/math] has no meaning. I will appretiate if anyone can remark me if I'm making or if I made a mistake in my logic(method) Thanks in advance, and excuse me if my coding is wrond.