Forgetting for a moment that the equation $e^{\sin x} - e^{-\sin x} - 4 = 0$ has no real solution, if we approach the final equation (which we get after solving the above equation)
$e^{\sin x} = 2 \pm \sqrt 5$
and do log on both sides then we get
$\log e^{\sin x} = \log (2 \pm \sqrt 5)$
$\implies \sin x = \log(2 + \sqrt 5)$ (as $\log (2 - \sqrt 5$ is a complex number)
$\implies \sin x = 0.626962921 \in [-1,1]$
Thus $x$ ought to have a real root on the above value, but it doesn't.
What is wrong with this approach besides not getting required result ?
For a start, $e^{\sin x} + e^{-\sin x} - 4 = 0$ yields the quadratic equation:
$$z^2 - 4 z + 1 = 0$$ where $z = e^{\sin x}$.
This yields: $$z = \dfrac {-(-4) \pm \sqrt {(-4)^2 - 4 \times 1 \times 1} } {2 \times 1} = 2 \pm \sqrt 3$$
This gives $2 + \sqrt 3 \approx 3.73$ and $2 - \sqrt 3 \approx 0.26$
Taking natural logarithms of this gives $$\sin x = \ln (2 + \sqrt 3) \approx 1.31, \sin x = \ln (2 - \sqrt 3) \approx -1.31$$
neither of which yields a real solution for $\sin x$.
Your biggest mistake is taking the logarithm base 10 when you should be taking the logarithm base $e$.
EDIT: I note that the equation in the title does show $e^{\sin x} - e^{-\sin x} - 4 = 0$ which is different from what is in the body of the question.
It is this sort of carelessness which you need to guard against.
In this case, yes, you are correct, $z = 2 \pm \sqrt 5$.
But then: $$\sin x = \ln (2 + \sqrt 5) \approx 1.44$$
which again has no real solution in $x$.