# Number of solutions of a trigonometric equation depending on the value of n

Question: For each integral $n$, find the number of solutions of $2x=(2n+1)×{\pi}(1-\cos x)$

I tried to solve first by simplifying then by using graphs of known equations. On simplifying I got $$\frac{x}{(2n+1)×{\pi}} = \sin^2 \frac{x}{2}$$ Then the right hand side (if plotted separately) of the equation is a straight line with slope $$\frac{1}{(2n+1)×{\pi}}$$. The LHS has the following graph:-

My idea is that the solutions will be at the intersection of the graph of LHS & RHS. On checking cases for $n=1$ i found there are 3 solutions , for $n=2$ i found there are 5 solutions .

My claim is that for every n there are $2n+1$ solutions, since by observation there is there is one intersection of the graph of LHS and RHS has exactly $1$ intersection between $x= (k-1)\pi$ and $k\pi$ for all $k \leq 2n+1$ ,an integer. But answer is given exactly $2n+3$ solutions. Is my claim wrong or it has any flaw? Please help. Thanks in advance.

• Please read this tutorial on how typeset mathematics on this site. Also, in the last paragraph, you appeared to write $(k - 1) \times /\pi$. What did you actually mean? May 7, 2017 at 11:23
• My question is 2x=(2n+1)×pi(1-cosx) May 7, 2017 at 17:50
• @N.F.Taussig for example n=1 we get a solution between \pi and 2\pi as shown in graph. May 8, 2017 at 13:52
• Moreover my question is why can't we solve graphically? May 8, 2017 at 13:57

i would write the equation in the form $$2x+\frac{2n+1}{\pi}\cos(x)-\frac{2n+1}{\pi}=0$$ when we define $$f(x)=2x+\frac{2n+1}{\pi}\cos(x)-\frac{2n+1}{\pi}$$ then we get $$f'(x)=0$$ if $$\sin(x)=\frac{2\pi}{2n+1}$$ and we have that $$|\sin(x)|\le 1$$