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
 A: 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. 
A: $g^{-1}(1)$ is the value $a$ such that $g(a) = 1$. So we're looking for a value that comes out to a nice round number when we plug it into $g$.
The functions $\sin x$ and $\cos x$ have "nice" values at integer multiples of $\frac{\pi}{2}$ and $e^x = 1$ if $x=0$, so maybe try some of those possibilities.
A: There's one "obvious" answer: try some small numbers whose sine and cosine you know.  But the function is not one-to-one on $\mathbb R$, and there are infinitely many other real values $x$, not expressible in "closed form", for which $g(x) = 1$.
A: Hint-You just need to guess for real $x$ for which $e^{-x}sinx$ is zero and $cosx=-1$. As $e^{-x}$ is not zero for finite real $x$, your work is reduced to find $x$ for which $sinx=0$ and $cosx=-1$.
A: Hint:  Show that each interval $[2\pi n,2\pi(n+1)]$ (with $n\in\mathbb{Z}$) contains exactly two values of $x$ for which $e^{-x}\sin x-\cos x=1$  by considering the two possibilities $\sin x=0$ and $\sin x\not=0$ separately.  When $\sin x=0$, you should be able to find the value of $x$ precisely.  When $\sin x\not=0$, draw a sketch of the two curves
$$y=e^{-x}\quad\text{and}\quad y={1+\cos x\over\sin x}$$
Additional hint (if you get stuck sketching the curve $y=(1+\cos x)/\sin x$):

 $${1+\cos x\over\sin x}=\cot(x/2)$$ and the cotangent decreases monotonically from $+\infty$ to $-\infty$ over and over again.

A: See the graph of the function and you ca easily see the answer:

A: See the graph. g (x) is period and hence inverse does not exist. So you are left with numerical analysis..
