If $f(1-x)-2f(x) = 1-x$, then solve for $f(\sin x-\cos x) = \frac{\sqrt{2}-4}{6}.$ 
If $f(1-x)-2f(x) = 1-x$, then solve for 
  $$f(\sin x-\cos x) = \frac{\sqrt{2}-4}{6}.$$           

How do I do this? I'm a bit confused with notations function notations; what's the difference between $f(x), f(x+2), f(2x)$? etc... 
For example if $f(x)= x+1$, then $f(x+1)= (x+1)+1$, right? $f(x)$ is the base function where $x$ is the input; $f(x+1)$ is the same function machine but with $x+1$ as the input.
Could you please elaborate... :D
sorry for the bad phrasing...

(1)$$f(1-\color{red}{x})-2f(\color{red}{x})=1-\color{red}{x} $$ 
$$f(1-\color{red}{(1-x)})-2f(\color{red}{1-x})=1-\color{red}{(1-x)}$$
Then 
(2) $$f(x)-2f(1-x)=x $$ 
$(2 \cdot (1)+(2)) \Rightarrow$
$$-3f(x)=2-x$$
$$f(x)=\frac{x-2}{3}$$
Hince, $$f(\sin x-\cos x) =\frac{\sin x-\cos x-2}{3}= \frac{\sqrt{2}-4}{6}.$$
$$\sin x - \cos x=\frac {\sqrt2}2$$
