$e^{Sin x}-e^{-Sinx}-4=0$ has no real roots 
Prove that $$e^{\sin x}-e^{-\sin x}-4=0$$ has no real roots.

My steps:
$$T=e^{\sin x}-e^{-\sin x}-4=0$$ 
$$T'=\cos x(e^{\sin x}+e^{-\sin x})=0$$ 
At $π/2$ and $3π/2$, $T'$ vanishes hence it should have at least one root. I agree that it could be found using the quadratic equation, but I would like to use my method. Even at $x=π/2$, $T>0$, at $x=0$, $T <0$. Any help would be appreciated.
 A: Observe that $e^{\sin x} - e^{-\sin x} -4 < e^{\sin x}  -4< e-4< 0$ for any real $x$. Hence, no real root exists.
A: $$\begin{array}{rcl}
e^{\sin x}-e^{-\sin x}-4 &=& 0 \\
e^{2\sin x}-1-4e^{\sin x} &=& 0 \\
e^{2\sin x}-4e^{\sin x}-1 &=& 0 \\
e^{\sin x} &=& \dfrac{4\pm\sqrt{4^2+4}}2 \\
e^{\sin x} &=& 2\pm\sqrt{5} \\
\sin x &=& \ln\left(2+\sqrt{5}\right) \\
\sin x &>& 1 \\
\end{array}$$
Desmos:

Just because the derivative vanishes at two points doesn't mean there's a root between.
A: One way is to see that it is equivalent to have $y-1/y-4=0$ where $y=e^{\sin(x)}$. This is the same as $y^2-4y-1=0$ so $y=\frac{4 \pm \sqrt{16+4}}{2}=\frac{4 \pm \sqrt{20}}{2}=2 \pm \sqrt{5}$. Are either of these in the range $[1/e,e]$? If so then they are outside the range of possible values of $y$.
As for your reasoning, in fact Rolle's theorem goes the other way around: if the derivative vanishes at most $k$ times in an open interval then there are at most $k+1$ roots in the corresponding closed interval, but there could be any smaller number of roots.
A: Take $e^{\sin x}=a $so 
$$e^{\sin x}-e^{-\sin x}-4=0 \to a-\frac{1}{a}-4=0\\$$multiply by $a$ so 
$$a^2-1-4a=0 \to (a-2)^2-4-1=0\\(a-2)^2=5 \\ a-2=\pm \sqrt5 \\a=2+\sqrt5 \\a=2-\sqrt 5$$
so 
$$e^{\sin x}=2+\sqrt 5 \to \sin x=ln(2+\sqrt 5) >ln(e)=1 \to  \nexists  x \in\mathbb{R}$$
$$e^{\sin x}=2-\sqrt 5 <0 \to a^u\text{ always} >0 \to  \nexists  x \in\mathbb{R}$$
A: Let's write $a=\sin(x)\in[-1,1]$, the equation is simply : $\quad\sinh(a)=2$
Yet $\sinh$ is strictly $\nearrow$ and $\sinh(1)\simeq 1.1752<2$, so there can't be solutions.
