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Algebra Difficulty 3.6 AMC 10/12 Find the answer

For each integer n>1n>1, let F(n)F(n) be the number of solutions to the equation sin⁡x=sin⁡(nx)\sin{x}=\sin{(nx)} on the interval [0,π][0,\pi]. What is ∑n=22007F(n)\sum_{n=2}^{2007} F(n)?

Pick one

Solution

F(2)=3F(2)=3
By looking at various graphs, we obtain that, for most of the graphs
F(n)=n+1F(n) = n + 1
Notice that the solutions are basically reflections across x=π2x = \frac{\pi}{2}.
However, when n≡1(mod4)n \equiv 1 \pmod{4}, the middle apex of the sine curve touches the sine curve at the top only one time (instead of two reflected points), so we get here F(n)=nF(n) = n.
3+4+5+5+7+8+9+9+⋯+20083+4+5+5+7+8+9+9+\cdots+2008
=(1+2+3+4+5+⋯+2008)−3−501= (1+2+3+4+5+\cdots+2008) - 3 - 501
=(2008)(2009)2−504=2016532= \frac{(2008)(2009)}{2} - 504 = 2016532 (D)\mathrm{(D)}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.