Range of $f(x)=\frac{\sin x -1}{\sqrt{3-2\cos x-2\sin x}}$ for a specified domain We are asked to find the range of the function $$f(x)=\frac{\sin x -1}{\sqrt{3-2\cos x-2\sin x}}, \;\;\text{for}\;0\le x\le2\pi$$
I tried to find the range of each basic function of cos and sin then see if I can use properties of inequality and algebraic combination of the basic functions to find the range of $f(x)$, but it turns out it is not that straightforward. It is not very hard to see that $f(x)\le0$, but I failed to find its minimum value. Then I tried several values of $x$ and guess that the range is [-1,0]. I have little clue how to find the range "rigorously".
Is there any quicker or smarter way to find the range of this kind of functions? (without calculus?)
Many thanks for any help!
 A: Since $\sin\frac\pi2=1$, we have $f(\frac{\pi}2)=0$, so it's clear that $0$ is in the range. You already got that $f(x)\leq 0$ for all $x$. The function is continuous on its domain, so we know that its range is $[-a,0]$ for some $a$.
Finding that minimum value is trickier. By plugging in values or graphing, you quickly obtain $-1$ as a guess, and then you can try to prove that $f(x)\geq -1$ for all $x$:
$\begin{align}
f(x)\geq -1 &\Leftrightarrow \frac{1-\sin x}{\sqrt{3-2\cos x-2\sin x}} \leq 1 \\
&\Leftrightarrow 1-\sin x\leq \sqrt{3-2\cos x-2\sin x} \\
&\Leftrightarrow 1-2\sin x+\sin^2x \leq 3-2\cos x-2\sin x \\
&\Leftrightarrow \sin^2x\leq 2(1-\cos x) \\
&\Leftrightarrow 1-\cos^2x \leq 2(1-\cos x) \\
&\Leftrightarrow \frac{1-\cos^2x}{1-\cos x} \leq 2 \\
&\Leftrightarrow 1+\cos x\leq 2
\end{align}$
That last inequality is certainly true, because cosine is never greater than $1$.
This doesn't really provide an elegant way to obtain the guess $-1$ in the first place. Calculus is great for that, it turns out. At any rate, since $f(0)=-1$, we now know the range is $[-1,0]$.

Post edited: my previous calculation contained an error; it should be correct now.
A: Let $-\sqrt y=\dfrac{\sin x-1}{\sqrt{3-2\sin x-2\cos x}}$ as the later is $\le0$ where $y\ge0$
Squaring both sides $$(\sin x-1)^2=y(3-2\sin x-2\cos x)$$
$$\iff\sin^2x+2(y-1)\sin x+1+y(2\cos x-3)=0$$
As $\sin x$ is real, the discriminant must be $\ge0$
i.e., $$(y-1)^2\ge1+y(2\cos x-3)$$
$$\iff y\{y-(2\cos x-1)\}\ge0$$
As $y\ge0,$ this $\implies y\ge$max$(0,2\cos x-1)=1$ if $x=2n\pi$
$$\implies\sqrt y\le1\iff-\sqrt y\ge-1$$
