Find the range of $f(x)=\frac1{1-2\sin x}$ Question:

Find the range for $f(x)= 1/(1-2\sin x)$

Answer :

$ 1-2\sin x \ne 0 $ 
$ \sin x \ne 1/2 $

My approach:

For range : 
$ -1 ≤ \sin x ≤ 1 $ 
$ -1 ≤ \sin x < 1/2$   and   $1/2<\sin x≤1 $  , because $\sin x≠1/2$ 
$ -2 ≤ 2\sin x <1 $ and $ 1< 2\sin x≤2 $ 
$ -1 < -2\sin x ≤2 $ and $ -2≤ -2\sin x<-1 $ 
$ 0 < 1-2\sin x ≤3   $ and $ -1≤ 1-2\sin x<0 $ 

I am stuck at the last step. If I take reciprocal i.e., $\frac 1{1-2\sin x}$ I get :
$ \frac10 < \frac1{1-2\sin x} ≤\frac13   $ and $ -1≤ \frac1{1-2\sin x} < \frac 10 $ 
$1/0$ can be interpreted as infinity so the second equation gives range of $f(x)$ as $[-1,∞)$. But the correct answer is : $(−∞,−1]∪[\frac13,∞)$
Any alternative solutions are welcome :)
P.S. : Also $\frac1{1-2\cos x}$ will also have the same range, right? As $\sin x$ and $\cos x$ both lie between $[-1,1]$. So the answer will proceed in similar fashion to the given one.


because $\sin x\neq 1/2$

That is not valid as it's circular reasoning, which is a fallacy in argument.
So, you had:
$-1\leq\sin x \leq1$
$\Rightarrow-2\leq2\sin x \leq2$
$\Rightarrow2\geq-2\sin x \geq-2$, that is, $-2\leq-2\sin x \leq2$
$\Rightarrow-1\leq1-2\sin x \leq3$
Now, when we take reciprocal, similar to what we do when we multiply the whole inequality by "$-$" the inequality reverses. So:
$\Rightarrow\frac1{-1}\geq\frac1{1-2\sin x} \geq\frac13$
$\therefore\frac1{1-2\sin x}\geq\frac13 \ $ also, $ \ \frac1{1-2\sin x} \leq -1$
Thus, we get our range as $(-\infty,-1]\cup[\frac13,\infty)$
But, note that $1/0$ is not defined so that means $(1-2\sin x)$ shouldn't be $0$. 
So, $\sin x\neq \frac12$.

Yup, replacing $\sin x$ by $\cos x$ won't change the answer, it'll behave the same way.
