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So I am willing to shift the ratio between the positive and the negative area of $\sin(x)$ by adding q. So for instance I would like to say that the positive area is 2 times larger than the negative by shifting $\sin(x)$ $q$ units up.
So I wrote $\sin(x)+q$ but I wanted my function to go through $(0,0)$, therefore I will get $$f:\mathbb{R}\to\mathbb{R}$$$$ x\mapsto\sin(x-\arcsin(q))+q$$ with $f(x)=0$ for $x=2\pi n$ or $x=\pi(2n+1)+2arcsin(q)$.

Now lets say I want the positive area 2 times larger than the negative area. This will give me: $$\int_{0}^{p} f(x) \,dx = -2*\int_{p}^{2\pi} f(x) \,dx$$ where $p=\pi+2arcsin(q)$.
However when I tried to solve it I would get an equation which had a complex outcome so that must be wrong. Therefore I assume that I am tackling this problem from the wrong angle. Can anyone help me where I went wrong?

EDIT This has been solved.

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    $\begingroup$ You forgot a minus sign on the integral. You have just set a positive integral equal to a negative one. Also, the negative integral is wrong because the right end of the interval also ends up being positive. $\endgroup$ May 27, 2021 at 18:12
  • $\begingroup$ I indeed forgot the negative sign, but it doesn't end up positive since $f(2\pi)=0$ $\endgroup$
    – Bessel
    May 27, 2021 at 18:19
  • $\begingroup$ @NinadMunshi, Thank you for your help, I can't believe it was such a small mistake which had me poundering for hours. $\endgroup$
    – Bessel
    May 27, 2021 at 18:29

1 Answer 1

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I made a small mistake but it has now been solved. If you work the example out you will get $$qp+2\sqrt{(1-q^2)}-4q\pi=0 \implies q ≈ 0.217233628211222...$$

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