Please only give hints, not answer: if $g(x)=e^{-x}\sin x-\cos x$ what is $g^{-1}(1)$ question
$g(x)=e^{-x}*\sin x-\cos x$ 

what is $ g^{-1}(1)$

my steps
i know that
$1=e^{-x}\sin x-\cos x$
thus 
$\ln 1=-x\sin x-\cos x$

here is a possibility:
$\ln^21=(-x-2x\sin x\cos x)^2 $
and that is where i got stuck
Hint: $e^{−x}$ is difficult to evaluate for even simple values, so if you can find the root by hand, there must be some way to offset, or negate the effort of calculating $e^{−x}$. $\sin x$ must have a value that makes the value of $e^{−x}$ irrelevant.
See the graph. g (x) is period and hence inverse does not exist. So you are left with numerical analysis..