How can we solve this equation? $$e^{\sin x} + e^ {\cos x} = e + 1$$
I know that $x=0$ and $x=\pi/2$ are solutions, as well as their periods, and that $x$ has to be between $0$ and $\pi/2$, but I don't know how to prove there aren't any other solutions.
As suggested by @zhw, I'll place here a proof of the statement: the only real roots of the equation $$ e^{\sin x}+e^{\cos x}=e+1 $$ are $x=2\pi k,$ $x=\frac{\pi}{2}+2\pi k,$ $k\in\mathbb{Z}.$
From $2\pi-$periodicity of $\sin$ and $\cos$ it is sufficient to show that the only roots from $[0,2\pi)$ are $x=0$ and $x=\frac{\pi}{2}.$ For $\frac{\pi}{2}<x<2\pi$ one of functions $\sin$ or $\cos$ is negative, hence in the sum $e^{\sin x}+e^{\cos x}$ one summand is $<1.$ The other is always $\leq e$ and the value $e+1$ can't be obtained.
It remains to consider $x\in[0,\frac{\pi}{2}].$ Let $f(x)=e^{\sin x}+e^{\cos x},$ $0\leq x\leq \frac{\pi}{2}.$ Then for $0<x<\frac{\pi}{2}$ $$ f'(x)=\cos xe^{\sin x}-\sin xe^{\cos x}=\sin x\cos x\bigg(\frac{e^{\sin x}}{\sin x}-\frac{e^{\cos x}}{\cos x}\bigg)=\sin x\cos x (g(\sin x)-g(\cos x)) $$ where $g(t)=\frac{e^t}{t}.$ For $0<t<1$ the function $g$ is decreasing, as $g'(t)=\frac{e^t(t-1)}{t^2}.$ Then $$ f'(x)>0\Leftrightarrow g(\sin x)>g(\cos x)\Leftrightarrow \sin x<\cos x $$ Function $f$ is strictly decreasing on $[0,\frac{\pi}{4}]$ and strictly increasing on $[\frac{\pi}{4},\frac{\pi}{2}].$ As $e+1=f(0)=f(\frac{\pi}{2}),$ for all $x\in(0,\frac{\pi}{2})$ we have $f(x)<e+1.$