This equation was part of a bigger calc question for a weekly assignment.
$$\frac{4}{\pi}\arctan{(1 + x)}\cosh{x} = 1$$
I found the solution to be $x = 0$ by inspection, but was wondering if this is solvable through algebraic methods, or anything other than numerical (graphing etc).
I certainly was not able to myself but maybe someone else can teach me something new?
Set $f(x) := \arctan(1+x)\cosh(x)$ for $x \in \mathbb{R}$.
If $x \le -1$, then $f(x)<0$, so $x$ is not solution.
On $[0,+\infty)$, $f$ is increasing function as the product of two non-negative increasing functions. We can check by derivation that $f$ is also increasing on $[-1,0]$.
Indeed, for every $x \in \mathbb{R}$,
$$f'(x) = \frac{1}{1+(1+x)^2}\cosh(x)+\arctan(1+x)\sinh(x),$$ so $f'(x)$ a the same sign as $$f'(x) = \frac{1}{1+(1+x)^2}+\arctan(1+x)\tanh(x).$$ Now assume $x \in [-1,0]$, so we set $x=-t$ with $t \in [0,1]$. Then $f'(x)$ a the same sign as $$\frac{1}{1+(1-t)^2}-\arctan(1-t)\tanh(t).$$ In the one hand, $1 \le 1+(1-t)^2 \le 2$, so $$\frac{1}{1+(1-t)^2} \ge \frac{1}{2}$$ In the other hand, convexity inequalities $\arctan(1-t) \le 1-t$ and $\tanh(t) \le t$ yield $$\arctan(1-t)\tanh(t) \le (1-t)t \le \frac{1}{4}.$$ Hence $$\frac{1}{1+(1-t)^2}-\arctan(1-t)\tanh(t) \ge \frac{1}{4}.$$ so $f'(x)>0$.
Therefore, since $f(0)=\pi/4$, $0$ is the only solution of your equation.