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I need to find all the solutions to the equation $$\sin x=ax$$ where $a \in \mathbb{R}$.

If $\vert a \vert \geq 1$, I know that the only solution is $x=0$. My problem is when $\vert a \vert \leq 1$ because the graph of $ax$ intersects the graph of $\sin x$ in more points than the origin. I tried with Lagrange multipliers but I don't think that is the correct way. Any suggestions?

Aritra Das
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  • You could be interested by http://math.stackexchange.com/questions/976462/a-1-400-years-old-approximation-to-the-sine-function-by-mahabhaskariya-of-bhaska and http://math.stackexchange.com/questions/871516/how-could-i-improve-this-approximation – Claude Leibovici May 04 '16 at 07:49
  • Lagrange Multipliers? For what? And I don't think it's analytically possible to find out all the solutions, though the number of solutions or intervals they lie in can be calculated. To find the exact roots, either try graphical methods or numerical methods. – Aritra Das May 04 '16 at 07:49
  • You could also use the approximation in Claude's first link for $\sin x$ by equating to $a(x+2n\pi)$. The $2n$ is because the approximation is for $[0,\pi]$ and to find all solutions, you need the other cases as well. Since period of $\sin x$ is $2\pi$, equating $\sin x = ax$ is the same as $\sin (x-2n\pi)=ax$ and put $x-2n\pi\to t$ – Aritra Das May 04 '16 at 07:59

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