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. Accord