# Showing $2x=\left( 2n+1\right) \pi \left( 1-\cos x\right)$ has $2n+3$ roots when $n\in \mathbb{Z}_+$

I am struggling to show that the equation $$2x=\left( 2n+1\right) \pi \left( 1-\cos x\right)$$ where n is a positive integer, has $2n+3$ roots and no more and also if it possible to indicate their locations (even roughly) ?

I am unsure how proceed with this one due to the presence of both $\cos x$ and $2x$ and would be great full if some one could give me a hint.

Cheers

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Draw a picture!

One wants to enumerate the intersections of the graph of the function $f:x\mapsto1-\cos(x)$ with the straight line $x\mapsto2x/((2n+1)\pi)$. Since $0\leqslant f\leqslant2$, these points are all in $[0,(2n+1)\pi]$. One of these is $0$, two of these are in the interval $I_k=(2k\pi-2\pi,2k\pi)$, for each $1\leqslant k\leqslant n$, and two are in the interval $(2n\pi,(2n+1)\pi]$ (one of those being $(2n+1)\pi$).

To show this, consider the function $g:x\mapsto f(x)-2x/((2n+1)\pi)$. Then $g(2k\pi-2\pi)\lt0$, $g(2k\pi-\pi)\gt0$, $g(2k\pi)\lt0$, and $g'(x)=\sin(x)-2/((2n+1)\pi)$ hence $g'$ is negative, then positive, then negative on $I_k$. This shows that $g$ has exactly two zeroes on $I_k$, one in $(2k\pi-2\pi,2k\pi-\pi)$ and the other in $(2k\pi-\pi,2k\pi)$.

Finally, there are exactly $2n+3$ intersection points.

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Hint: The equation easily transforms to $x=(2n+1)\frac{\pi}{2}\sin^{2}x$. It is easy to see that there are no roots for $x<0$ and $x>(2n+1)\frac{\pi}2$. Also, $(2n+1)\frac{\pi}{2}\sin^{2}x=0$ at $x= k\pi$.

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A downvote without comments does not really help me improve my answer. –  Shitikanth Apr 14 '12 at 9:41