Let's get a little trigy

The given function can be written as

$f(x) = \dfrac{ \sin(x) - 1 }{ \sqrt{ (\cos(x) - 1 )^2 + (\sin(x) - 1 )^2 } }$

which is exactly the sine of the angle that the vector from the point $(1,1)$ to the point $(\cos(x) ,\sin(x))$ makes with the positive x-axis. This angle is clearly always in the third quadrant, so its sine is between $-1$ and $0$ .

Thanks for the idea from @Hosam Hajjir , my solution uses half-angle tangent substitution .

Given that $f(x) = \dfrac {\sin x-1}{\sqrt{3-2\cos x - 2 \sin x}} = \dfrac {\sin x -1}{\sqrt{3-2\sqrt 2\sin \left(2+\frac \pi 4\right)}}$ . Note that $g(x) = \sin x - 1 \in [-2, 0]$ and $h(x) = \sqrt{3-2\sqrt 2\sin \left(2+\frac \pi 4\right)} \in [\sqrt{3-2\sqrt 2}, \sqrt{3+2\sqrt 2}]$ . That is $g(x) \le 0$ and $h(x) > 0$ $\implies f(x) \le 0$ .

Now consider the lower limit of $f(x)$ as follows:

$\begin{aligned} f(x) & = \frac {\sin x-1}{\sqrt{3-2\cos x - 2 \sin x}} \\ & = \frac {-(1-\sin x)}{\sqrt{(1-\cos x)^2 +(1-\sin x)^2}} & \small \color{#3D99F6} \text{For }\sin x - 1 \ne 0 \\ & = \frac {-1}{\sqrt{\left(\frac {1-\cos x}{1-\sin x}\right)^2 + 1}} & \small \color{#3D99F6} \text{Let }t = \tan \frac x2 \implies \sin x = \frac {2t}{1+t^2} \\ & = \frac {-1}{\sqrt{\frac {4 t^4}{(t-1)^4} + 1}} & \small \color{#3D99F6} \text{and } \cos x = \frac {1-t^2}{1+t^2} \\ \implies f(x) & \ge - 1 & \small \color{#3D99F6} \text{Equality occurs when }t = 0 \text{ or }x = 0,2\pi \end{aligned}$

Therefore, $f(x) \in \boxed{[-1,0]}$ .