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Does all values of $x$ in $\mathbb R$ satisfy equation: $$e^{2\ln(\sin(x))} = 1 - e^{2\ln(\cos(x))}$$

I am asking this, because by checking WolframAlpha solution there is an answer: (all values for $x$ are solutions over reals), but we know that $\ln(0)$ is undefined, same for negative numbers.

Wolfram Alpha solution

Therefore I assume in $\mathbb R$, zero and negative numbers doesn't satisfy this equation.

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  • $\begingroup$ What exactly do you mean by $\ln*\sin(x)$? Does it refer to $\ln(\sin(x))$ or what is the meaning of this notation? $\endgroup$ – mrtaurho Mar 13 at 17:52
  • $\begingroup$ sorry, my bad. Yes. Will fix equation $\endgroup$ – Peter Parada Mar 13 at 17:53
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    $\begingroup$ Wolfram Alpha will always give explicitly Wrong Answer for some input. Any real mathematician will always be able to easily find a limit that exists but WA says it doesn't, or vice versa. Whenever someone reports a wrong answer, the programmers just hack a 'fix' based on random guesswork rather than mathematics. Bottom line: Don't trust WA to be correct even on very simple questions. $\endgroup$ – user21820 Mar 27 at 6:34
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$$1=e^{2\ln\sin(x)} +e^{2\ln \cos(x)}= e^{\ln\sin^2(x)}+e^{\ln \cos^2(x)} = $$

$$ = \sin^2(x)+\cos^2(x) = 1 $$

So this is true for all $x$ such that $\sin x>0$ and $\cos x>0$ (because then is $\ln $ defined).

So $$\boxed{x\in \color{red}{\bigcup_{k\in \mathbb{Z}} (0+2\pi k,{\pi\over 2}+2\pi k)}}$$

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  • $\begingroup$ Great. I was thinking the same. Does that mean solution on Wolfram is wrong? $\endgroup$ – Peter Parada Mar 13 at 18:04
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    $\begingroup$ (+1) For myself I totally forgot to think about the existence of the natural logarithm as it is only assured for $\sin(x),\cos(x)>0$. $\endgroup$ – mrtaurho Mar 13 at 18:04
  • $\begingroup$ @PeterParada I'm not sure what is Wolfram doing. It draw you a whole line for $f(x) =e^{\ln x}$ which is absurd to me. $\endgroup$ – Aqua Mar 13 at 18:07
  • $\begingroup$ Thank you. I will write them for further clarification. $\endgroup$ – Peter Parada Mar 13 at 18:07
  • $\begingroup$ @PeterParada: I think Wolfram implicitly assumes that $x$ is in the domain; you might need to check documentation to see if this is standard. $\endgroup$ – Clayton Mar 13 at 18:08
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Note that : $$e^{2\ln \sin x} + e^{2 \ln \cos x} = 1 \Rightarrow e^{\ln (\sin x)^2} + e^{\ln (\cos x)^2} = 1 \Leftrightarrow \sin^2x + \cos^2x = 1 \rightarrow \text{true} \; \forall x \in \mathbb R$$ Restrictions apply so as the initial expression holds, so that narrows down the solution set. Note the usage of $\Rightarrow$ instead of an $\Leftrightarrow$ at the start.

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