Solve for the following: $\frac{\pi}{4}=\frac{e^x-e^{-x}}{2}+\arctan(x+1)$

Question

Question

Solve for the following : $$\frac{\pi}{4}=\frac{e^x-e^{-x}}{2}+\arctan(x+1)$$

Clearly I would mutliply both sides by 2 but then I would get the following :

$$\frac{\pi}{2}=e^x-e^{-x}+\arctan(x+1)$$

I was wondering how I would simplify from here. Would i convert $\arctan$ into $\frac{\arcsin}{\arccos}?$

preferably done without using hyperbolic trig functions

Answer 1

You can guess that $0$ is a solution, and because the derivative of $e^x-e^{-x}+2\arctan(x+1)$ is always positive, we know that $0$ is the only solution.

Answer 2

Let $f(x)=\frac{e^x-e^{-x}}{2}+\arctan(x+1)$.

Since $f'(x)=\frac{e^x+e^{-x}}{2}+\frac{1}{x^2+2x+2}>0$,

we get that $f$ is an increasing function.

Thus, our equation has one root maximum and since $0$ is a root, we get an answer: $\{0\}$.

I fixed a typo.