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For our homework, our teacher assigned us an extremely demanding problem.$$\sin(-x)+\cot(x)=1$$ This problem may seem easy but the answer is incredibly complex. There are two sets of (real) solutions, here they are: $$x=2\left(\pi n+\arctan\left(\frac{1}{3}\left(-1-\frac{4\cdot2^{\frac{2}{3}}}{\sqrt[3]{13+3\sqrt{33}}}+\sqrt[3]{2\left(13+3\sqrt{33}\right)}\right)\right)\right),n\in \mathbb{Z},\tan(x)\neq0$$ and $$x=\frac{1}{2}\left(4\pi n-\pi\right),n\in \mathbb{Z}$$ I have no idea how to attain either of these solutions. If someone can help me get either of these solutions with clear steps, that would be incredibly helpful.

Blue
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SovietWizard
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    At least you should be able to find the second set of solutions? – Ma Ye Nov 21 '24 at 03:09
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    Try pulling out the negative sign from the sine function and square both sides. You should be able to get a cubic equation in terms of $\cos$ –  Nov 21 '24 at 03:29
  • Okay thanks, I did that and it doesn't work. For all of our work we aren't allowed to use a calculator or calculus, and there is an extraneous solution where that above complex solution is written in terms of arccos, and then you know I get something like plus minus arccos(...) + 2pin, it's very heavy to prove that extraneous solution is false though without calculus, at least the way I did. – SovietWizard Dec 17 '24 at 05:53

1 Answers1

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Hint

I suppose that the first formula came from the tangent half-angle substitution

$$f(x)=\cot(x)-\sin(x)-1$$

$$x=2\tan^{-1}(t) \quad \implies \quad g(t)=\frac{(t+1) \left(t^3+t^2+3 t-1\right)}{2 t \left(t^2+1\right)}$$

Using the hyperbolic solution, the real root of the cubic is just $$t=\frac{4\sqrt{2}}{3} \sinh \left(\frac{1}{3} \sinh ^{-1}\left(\frac{13}{8 \sqrt{2}}\right)\right)-\frac{1}{3}$$ This looks (at least to me) abit nicer.

Claude Leibovici
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