
Equation
HI everybody,
I must show that the following equation: cosx+cos2x+...+cosnx=0 $\displaystyle (n\in \mathbb{N}*)$ accepts at least one solution in $\displaystyle [0,\pi]$.
1)We know that f(x)=cosx+cos2x+...+cosnx is continious in $\displaystyle [0,\pi]$.
2)So i must show that $\displaystyle f(0)*f(\pi)<0$, BUT HOW?
CAN YOU HELP ME PLEASE?
AND THANKS ANYWAY.

For $\displaystyle f(0)\times f(\pi)$ consider
$\displaystyle f(\pi)= \cos(\pi)+\cos(2\pi)+\cos(3\pi)+\cos(4\pi)+\dots$
$\displaystyle f(\pi)= 1+11+1+\dots$
and
$\displaystyle f(0) = \cos(0)+\cos(2\times 0 )+\cos(3\times 0)+\cos(4\times 0)+\dots$
$\displaystyle f(0) = 0+0+0+0+\dots$
Can you see a pattern? What are the implications for n being either even or odd for $\displaystyle f(\pi)$ ?