# Equation with inverse trigonometric functions and logarithms

Solve over reals

$$\log_\frac{\pi}{2}\left(\arcsin\, \{x\}\right)+\log_\frac{\pi}{2}\left(\arccos\,\{x\}\right)=\frac{2}{\log_\frac{\pi}{4}\left(\arctan e^{\lfloor x\rfloor} + \operatorname{arccot} e^{\lfloor x\rfloor}\right)}$$

where $$\{x\}$$ is the fractional part of $$x$$ and $$\lfloor x\rfloor$$ the floor function.

For the left side I have

$$\log_\frac{\pi}{2}\left(\arcsin\, \{x\}\right)+\log_\frac{\pi}{2}\left(\arccos\,\{x\}\right)=\log_\frac{\pi}{2}\left(\arcsin\, \{x\}\cdot \arccos\,\{x\}\right)$$

but I don't know what to do with the right side and I don't know how to proceed.

• What is the relationship between $\arctan x$ and $\operatorname{arccot} x$ for all real $x$?
– dfnu
Commented Feb 18, 2020 at 21:01
• @dnfu. I found online that $\arctan x +\operatorname{arccot} x = \frac{\pi}{2}$. But still, I am stuck at $\log_\frac{\pi}{2}\left(\arcsin\, \{x\}\cdot \arccos\,\{x\}\right) = \frac{2}{\log_\frac{\pi}{4} \frac{\pi}{2}}$. Also, will this question be closed? (I don't know what I did wrong).
– user750196
Commented Feb 18, 2020 at 21:35
• Recall that $0\leq \{x\} < 1$. Considering this, what is the relationship between $\arcsin \{ x\}$ and $\arccos \{ x\}$? Also $\log_a b = \frac1{\log_b a}$.
– dfnu
Commented Feb 18, 2020 at 21:57
• I have no idea why the question has been downvoted, by the way.
– dfnu
Commented Feb 18, 2020 at 22:06

We have $$\arccos\{x\} = \frac{\pi}2 - \arcsin\{x\}$$ and $$\arctan e^{\lfloor x \rfloor}+\operatorname{arccot} e^{\lfloor x \rfloor}=\frac{\pi}2.$$
Therefore your equation is equivalent to $$\log_{\pi/2}\left[\arcsin\{x\}\left(\frac{\pi}2-\arcsin\{x\}\right)\right]=\frac{2}{\log_{\pi/4}\frac{\pi}2}.$$ Recall now that $$\log_a b\cdot \log_b a =1$$ and write then $$\log_{\pi/2}\left[\arcsin\{x\}\left(\frac{\pi}2-\arcsin\{x\}\right)\right]=\log_{\pi/2}\left(\frac{\pi}4\right)^2.$$ Equating the logarithm arguments yields then $$\arcsin\{x\}\left(\frac{\pi}2-\arcsin\{x\}\right)= \frac{\pi^2}{16}$$ Or equivalently $$\begin{eqnarray}\arcsin^2\{x\}-\frac{\pi}2\arcsin\{x\}+\frac{\pi^2}{16}&=&0\\ \left(\arcsin\{x\}-\frac{\pi}4\right)^2 &=& 0 \end{eqnarray}$$ which leads to $$\{x\} = \frac1{\sqrt 2},$$ i.e. $$x = k + \frac1{\sqrt 2}, \ \ k\in \Bbb Z.$$
• Thank you, I found something similar: $\log_\frac{\pi}{2}\left(\arcsin\, \{x\}\cdot \arccos\,\{x\}\right) = \log_\frac{\pi}{2} \left(\frac{\arcsin\, \{x\}+ \arccos\,\{x\}}{2}\right)^2$$\implies \arcsin\, \{x\}\cdot \arccos\,\{x\}= \left(\frac{\arcsin\, \{x\}+ \arccos\,\{x\}}{2}\right)^2 \implies$$ \left(\arcsin\, \{x\}- \arccos\,\{x\}\right)^2=0\implies \arcsin\, \{x\}= \arccos\,\{x\} \implies \{x\}=\frac{\sqrt{2}}{2}$