Equation with inverse trigonometric functions and logarithms

Question

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.

Answer 1

Here is how I would proceed.

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. $$x = k + \frac1{\sqrt 2}, \ \ k\in \Bbb Z.$$